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3.3: Solving Trigonometric Equations

Created by: CK-12

Learning Objectives

  • Use the fundamental identities to solve trigonometric equations.
  • Express trigonometric expressions in simplest form.
  • Solve trigonometric equations by factoring.
  • Solve trigonometric equations by using the Quadratic Formula.

By now we have seen trigonometric functions represented in many ways: Ratios between the side lengths of right triangles, as functions of coordinates as one travels along the unit circle and as abstract functions with graphs. Now it is time to make use of the properties of the trigonometric functions to gain knowledge of the connections between the functions themselves. The patterns of these connections can be applied to simplify trigonometric expressions and to solve trigonometric equations.

Simplifying Trigonometric Expressions

Example 1: Simplify the following expressions using the basic trigonometric identities:

a. \frac{1 + \tan^2 x}{\csc^2 x}

b. \frac{\sin^2 x + \tan^2 x + \cos^2 x}{\sec x}

c. \cos x - \cos^3x

Solution:

a.

\frac{1 + \tan^2 x}{\csc^2 x}&\ldots( 1 + \tan^2 x = \sec^2 x ) \text{Pythagorean Identity} \\\frac{\sec^2 x}{\csc^2 x} & \ldots (\sec^2 x  = \frac{1}{\cos^2 x}\ \text{and}\ \csc^2 x = \frac{1}{\sin^2 x})  \text{Reciprocal Identity} \\\frac{\frac{1}{\cos^2 x}}{\frac{1}{\sin^2 x}}  &= \left (\frac{1}{\cos^2 x} \right ) \div \left (\frac{1}{\sin^2 x} \right ) \\\left (\frac{1}{\cos^2 x} \right ) \cdot \left (\frac{\sin^2 x}{1} \right ) &= \frac{\sin^2 x}{\cos^2 x}\\& = \tan^2 x \rightarrow \text{Quotient Identity}

b.

 \frac{\sin^2 x + \tan^2 x + \cos^2 x}{\sec x} &\ldots (\sin^2 x + \cos^2 x = 1) \text{Pythagorean Identity} \\ \frac{1 + \tan^2 x}{\sec x} &  \ldots (1 + \tan^2 x = \sec^2 x) \text{Pythagorean Identity} \\\frac{\sec^2 x}{\sec x} & = \sec x

c.

& \cos x - \cos^3 x \\& \cos x (1 - \cos^2 x) \qquad \ldots  \text{Factor out}\ \cos x \ \text{and}\ \sin^2 x = 1 - \cos^2 x  \\& \cos x (\sin^2 x)

In the above examples, the given expressions were simplified by applying the patterns of the basic trigonometric identities. We can also apply the fundamental identities to trigonometric equations to solve for x. When solving trig equations, restrictions on x (or \theta) must be provided, or else there would be infinitely many possible answers (because of the periodicity of trig functions).

Solving Trigonometric Equations

Example 2: Without the use of technology, find all solutions \tan^2(x) = 3, such that 0 \le x \le 2 \pi.

Solution:

\tan^2 x & = 3 \\\sqrt{\tan^2 x} & = \sqrt{3} \\\tan x & = \pm \sqrt{3}

This means that there are four answers for x, because tangent is positive in the first and third quadrants and negative in the second and fourth. Combine that with the values that we know would generate \tan x = \sqrt{3} or \tan x = - \sqrt{3},x = \frac{\pi}{3},\frac{2 \pi}{3}, \frac{4 \pi}{3}, and \frac{5 \pi}{3}.

Example 3: Solve 2 \cos x \sin x - \cos x = 0 for all values of x between [0, 2\pi].

Solution:

& \qquad \ \cos x \ \ (2 \sin x -1) = 0 \rightarrow \text{set each factor equal to zero and solve them separately} \\& \qquad \quad  \downarrow \qquad \qquad  \searrow \\& \qquad \cos x = 0 \qquad \quad 2 \sin x = 1 \\& x = \frac{\pi}{2}\ \text{and}\ x = \frac{3\pi}{2} \quad \sin x = \frac{1}{2} \\& \qquad \qquad \qquad \qquad \quad x = \frac{\pi}{6}\ \text{and}\ x = \frac{5\pi}{6}

In the above examples, exact values were obtained for the solutions of the equations. These solutions were within the domain that was specified.

Example 4: Solve 2 \sin^2 x - \cos x - 1 = 0 for all values of x.

Solution: The equation now has two functions – sine and cosine. Study the equation carefully and decide in which function to rewrite the equation. \sin^2 x can be expressed in terms of cosine by manipulating the Pythagorean Identity, \sin^2 x + \cos^2 x = 1.

& \qquad \ \ 2 \sin^2 x - \cos x - 1 = 0 \\& 2 (1 - \cos^2 x) - \cos x - 1 = 0 \\& \quad \ \ 2 - 2 \cos^2 x - \cos x - 1 = 0 \\& \quad \  - 2 \cos^2 x - \cos x + 1 = 0 \\& \qquad \ \ 2 \cos^2 x + \cos x - 1 = 0 \\& \quad (2 \cos x -1)(\cos x + 1) = 0 \\& \qquad \swarrow \qquad \qquad \quad \ \searrow \\& 2 \cos x - 1 = 0 \quad \text{or}\quad \cos x + 1 = 0 \\& \cos x = \frac{1}{2} \qquad \qquad \quad \cos x = - 1 \\& x = \frac{\pi}{3} + 2 \pi k, k \epsilon Z \qquad x = \pi + 2 \pi k, k \epsilon Z \\& x = \frac{5\pi}{3} + 2 \pi k, k \epsilon Z

Solving Trigonometric Equations Using Factoring

Algebraic skills like factoring and substitution that are used to solve various equations are very useful when solving trigonometric equations. As with algebraic expressions, one must be careful to avoid dividing by zero during these maneuvers.

Example 5: Solve 2 \sin^2 x - 3 \sin x + 1 = 0 for 0 < x \le 2 \pi.

Solution:

& \quad 2 \sin^2 x - 3 \sin x + 1 = 0 \quad \text{Factor this like a quadratic equation} \\& (2 \sin x - 1)(\sin x - 1) = 0 \\& \qquad \ \downarrow \qquad \qquad \ \ \ \searrow \\& \ 2 \sin x - 1 = 0 \quad \text{or} \ \ \sin x - 1 = 0 \\& \quad \ \ \  2 \sin x = 1 \qquad \qquad \ \sin x = 1 \\& \qquad \ \ \sin x = \frac{1}{2} \qquad \quad \qquad \ \  x = \frac{\pi}{2}\\x & = \frac{\pi}{6} \ \text{and} \ x = \frac{5 \pi}{6}

Example 6: Solve 2 \tan x \sin x + 2 \sin x = \tan x + 1 for all values of x.

Solution:

Pull out \sin x

There is a common factor of (\tan x + 1)

Think of the -(\tan x + 1) as (-1)(\tan x + 1), which is why there is a -1 behind the 2 \sin x.

Example 7: Solve 2 \sin^2 x + 3 \sin x - 2 = 0 for all x, [0, \pi].

Solution:

& \quad 2 \sin^2 x +3 \sin x - 2 = 0 \rightarrow \text{Factor like a quadratic} \\& (2 \sin x -1)(\sin x + 2) = 0 \\& \quad \ \ \swarrow \qquad \qquad \quad \searrow \\& 2 \sin x - 1 = 0 \qquad \sin x + 2 = 0 \\& \qquad \ \sin x = \frac{1}{2} \qquad \quad \ \ \sin x = -2 \\x & = \frac{\pi}{6} \ \text{and} \ x = \frac{5 \pi}{6}\text{        There is no solution because the range of}\   \sin x\ \text{is}\ [-1, 1].

Some trigonometric equations have no solutions. This means that there is no replacement for the variable that will result in a true expression.

Example 8: Solve 4 \sin^3 x + 2 \sin^2 x - 2 \sin x - 1 = 0 for x in the interval [0, 2\pi].

Solution: Even though this does not look like a factoring problem, it is. We are going to use factoring by grouping, from Algebra II. First group together the first two terms and the last two terms. Then find the greatest common factor for each pair.

& \underbrace{ 4 \sin^3 x + 2 \sin^2 x } \qquad \underbrace{-2 \sin x -1} = 0 \\& 2 \sin^2 x {\color{blue}(2 \sin x + 1)} - 1{\color{blue}(2 \sin x + 1)}

Notice we have gone from four terms to two. These new two terms have a common factor of {\color{blue}2 \sin x + 1}. We can pull this common factor out and reduce our number of terms from two to one, comprised of two factors.

& \qquad 2 \sin^2 x {\color{blue}(2 \sin x + 1)} - 1 {\color{blue}(2 \sin x + 1) = 0} \\& \qquad \qquad \ \searrow \qquad \qquad \ \ \swarrow \\& {\color{blue}(2 \sin x + 1)}(2 \sin^2 x \ - \ 1) = 0

We can take this one step further because 2 \sin^2 x-1 can factor again.

{\color{blue}(2 \sin x + 1)} \left (\sqrt{2} \sin x - 1 \right ) \left (\sqrt{2} \sin x + 1 \right ) = 0

Set each factor equal to zero and solve.

2 \sin x + 1 & = 0 \qquad \quad \text{or}\qquad \quad && \sqrt{2} \sin x + 1 = 0\qquad \quad \text{or} \qquad \quad && \sqrt{2} \sin x - 1 = 0  \\2 \sin x & = -1 && \quad \ \ \sqrt{2} \sin x = -1 && \quad \ \ \sqrt{2} \sin x = 1 \\\sin x & = -\frac{1}{2} && \qquad \quad \sin x = - \frac{1}{\sqrt{2}} = - \frac{\sqrt{2}}{2} && \qquad \quad \sin x = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \\x & = \frac{7 \pi}{6}, \frac{11\pi}{6} && \qquad \qquad \ \ x = \frac{5\pi}{4},\frac{7\pi}{4} && \qquad \qquad \ \ x = \frac{\pi}{4},\frac{3\pi}{4}

Notice there are six solutions for x. Graphing the original function would show that the equation crosses the x-axis six times in the interval [0, 2\pi].

Solving Trigonometric Equations Using the Quadratic Formula

When solving quadratic equations that do not factor, the quadratic formula is often used. The same can be applied when solving trigonometric equations that do not factor. The values for a is the numerical coefficient of the function's squared term, b is the numerical coefficient of the function term that is to the first power and c is a constant. The formula will result in two answers and both will have to be evaluated within the designated interval.

Example 9: Solve 3\ \cot^2x - 3\ \cot x = 1 for exact values of x over the interval [0, 2\pi].

Solution:

3 \cot^2x - 3 \cot x&= 1  \\3 \cot^2x - 3 \cot x - 1 & = 0

The equation will not factor. Use the quadratic formula for \cot x, a =3, b = -3, c = -1.

\cot x & = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\\cot x & = \frac{- (-3) \pm \sqrt{(-3)^2 - 4(3)(-1)}}{2(3)}\\\cot x & = \frac{3 \pm \sqrt{9 + 12}}{6} \\\cot x & = \frac{3+ \sqrt{21}}{6} \qquad \qquad \qquad \quad \text{or} \quad && \cot x = \frac{3- \sqrt{21}}{6} \\\cot x & = \frac{3+4.5826}{6} && \cot x = \frac{3-4.5826}{6} \\\cot x & = 1.2638 && \cot x =-0.2638 \\\tan x & = \frac{1}{1.2638} && \tan x = \frac{1}{-0.2638} \\x & = 0.6694, 3.81099 && x = 1.8287, 4.9703

Example 10: Solve -5 \cos^2x + 9 \sin x + 3 = 0 for values of x over the interval [0, 2\pi].

Solution: Change \cos^2 x to 1 - \sin^2 x from the Pythagorean Identity.

-5 \cos^2x + 9 \sin x + 3 & = 0 \\ -5 (1 - \sin^2x) + 9 \sin x + 3 & = 0 \\-5 + 5 \sin^2x + 9 \sin x + 3 & = 0 \\5 \sin^2x + 9 \sin x - 2 & = 0

& \sin x = \frac{- 9 \pm \sqrt{9^2 - 4(5)(-2)}}{2(5)} \\& \sin x  = \frac{-9 \pm \sqrt{81+40}}{10} \\& \sin x = \frac{-9 \pm \sqrt{121}}{10} \\& \sin x = \frac{-9 + 11}{10} \ \text{and}\ \sin x = \frac{-9-11}{10} \\& \sin x = \frac{1}{5}\ \text{and}\ -2 \\&\sin^{-1} (0.2)\ \text{and}\ \sin^{-1}(-2)

x \approx .201\ rad and \pi -.201 \approx 2.941

This is the only solutions for x since -2 is not in the range of values.

To summarize, to solve a trigonometric equation, you can use the following techniques:

  1. Simplify expressions with the fundamental identities.
  2. Factor, pull out common factors, use factoring by grouping.
  3. The Quadratic Formula.
  4. Be aware of the intervals for x. Make sure your final answer is in the specified domain.

Points to Consider

  • Are there other methods for solving equations that can be adapted to solving trigonometric equations?
  • Will any of the trigonometric equations involve solving quadratic equations?
  • Is there a way to solve a trigonometric equation that will not factor?
  • Is substitution of a function with an identity a feasible approach to solving a trigonometric equation?

Review Questions

  1. Solve the equation \sin 2\theta = 0.6 for 0 \le \theta < 2 \pi.
  2. Solve the equation \cos^2 x = \frac{1}{16} over the interval [0, 2\pi]
  3. Solve the trigonometric equation \tan^2 x = 1 for all values of \theta such that 0 \le \theta \le 2\pi
  4. Solve the trigonometric equation 4 \sin x \cos x + 2 \cos x-2 \sin x - 1 = 0 such that 0 \le x < 2\pi.
  5. Solve \sin^2 x - 2 \sin x - 3 = 0 for x over [0, \pi].
  6. Solve \tan^2 x = 3 \tan x for x over [0, \pi].
  7. Find all the solutions for the trigonometric equation 2 \sin^2 \frac{x}{4}-3 \cos \frac{x}{4} = 0 over the interval [0, 2\pi).
  8. Solve the trigonometric equation 3 - 3 \sin^2 x = 8 \sin x over the interval [0, 2\pi].
  9. Solve 2 \sin x \tan x = \tan x + \sec x for all values of x\ \epsilon [0, 2\pi].
  10. Solve the trigonometric equation 2 \cos^2x + 3 \sin x - 3 = 0 over the interval [0, 2\pi].
  11. Solve \tan^2x + \tan x - 2 = 0 for values of x over the interval \left [- \frac{\pi}{2},\frac{\pi}{2} \right ].
  12. Solve the trigonometric equation such that 5 \cos^2 \theta - 6 \sin \theta = 0 over the interval [0, 2\pi].

Review Answers

  1. Because the problem deals with 2\theta, the domain values must be doubled, making the domain 0 \le 2\theta < 4\pi The reference angle is \alpha = \sin^{-1} 0.6 = 0.6435 2\theta = 0.6435, \pi - 0.6435, 2\pi + 0.6435, 3\pi - 0.6435 2\theta = 0.6435, 2.4980, 6.9266, 8.7812 The values for \theta are needed so the above values must be divided by 2. \theta = 0.3218, 1.2490, 3.4633, 4.3906 The results can readily be checked by graphing the function. The four results are reasonable since they are the only results indicated on the graph that satisfy \sin 2\theta = 0.6.
  2. \cos^2 x & = \frac{1}{16} \\\sqrt{\cos^2 x} & = \sqrt{\frac{1}{16}} \\\cos x & = \pm \frac{1}{4} \\\text{Then}\ \cos x & = \frac{1}{4} && \text{or} && \qquad \ \ \cos x = - \frac{1}{4} \\\cos^{-1} \frac{1}{4} & = x && && \cos^{-1} -\frac{1}{4} = x \\x  &= 1.3181\ radians && && \qquad \qquad \ x = 1.8235\ radians However, \cos x is also positive in the fourth quadrant, so the other possible solution for \cos x = \frac{1}{4} is 2\pi - 1.3181 = 4.9651\ radians and \cos x is also negative in the third quadrant, so the other possible solution for \cos x = - \frac{1}{4} is 2\pi - 1.8235 = 4.4597\ radians
  3. \tan^2 x & = 1 \\\tan x & = \pm \sqrt{1} \\\tan x & = \pm 1 so, \tan x = 1 or \tan x = -1. Therefore, x is all critical values corresponding with \frac{\pi}{4} within the interval. x = \frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}
  4. Use factoring by grouping. & 2 \sin x + 1 = 0 \quad \text{or} \qquad 2 \cos x - 1 = 0 \\& 2 \sin x = -1 \qquad \qquad \quad 2 \cos x = 1 \\& \ \ \sin x = - \frac{1}{2} \qquad \qquad \quad \cos x = \frac{1}{2} \\& \qquad \  x = \frac{7 \pi}{6}, \frac{11\pi}{6} \qquad \qquad \quad x = \frac{\pi}{3}, \frac{5\pi}{3}
  5. You can factor this one like a quadratic. & \quad \sin^2 x - 2 \sin x - 3 = 0 \\& \ (\sin x - 3)(\sin x + 1) = 0 \\& \sin x - 3 = 0 && \ \sin x + 1 = 0 \\& \quad \ \ \sin x = 3 \qquad \qquad \qquad \text{or} && \qquad \sin x = -1 \\& \qquad \quad \ x = \sin^{-1}(3) && \qquad \quad \ x = \frac{3 \pi}{2} For this problem the only solution is \frac{3\pi}{2} because sine cannot be 3 (it is not in the range).
  6. \tan^2 x &= 3 \tan x \\\tan^2 x - 3 \tan x &= 0 \\\tan x (\tan x - 3) &= 0 \\\tan x & = 0 \qquad \text{or} \qquad \tan x = 3 \\x & = 0, \pi  \qquad \qquad \quad \ \ x = 1.25
  7. 2 \sin^2 \frac{x}{4} - 3 \cos \frac{x}{4} = 0 & \quad 2 \left (1 - \cos^2 \frac{x}{4} \right ) - 3 \cos \frac{x}{4} = 0 \\& \qquad \ 2 - 2 \cos^2 \frac{x}{4} - 3 \cos \frac{x}{4} = 0 \\& \qquad \ 2 \cos^2 \frac{x}{4} + 3 \cos \frac{x}{4} - 2 = 0 \\& \left (2 \cos \frac{x}{4} - 1 \right ) \left (\cos \frac{x}{4} + 2 \right ) = 0 \\& \qquad \swarrow  \qquad \qquad \qquad \searrow\\& 2 \cos \frac{x}{4} - 1 = 0 \quad \text{or} \quad \cos \frac{x}{4} + 2 = 0 \\& \quad \ \ 2 \cos \frac{x}{4} = 1 \qquad \qquad \ \ \cos \frac{x}{4} = -2 \\& \qquad \cos \frac{x}{4} = \frac{1}{2} \\& \frac{x}{4} = \frac{\pi}{3} \quad \text{or} \quad \frac{5\pi}{3} \\& x = \frac{4 \pi}{3} \ \ \text{or} \quad \frac{20\pi}{3} \frac{20 \pi}{3} is eliminated as a solution because it is outside of the range and \cos \frac{x}{4} = -2 will not generate any solutions because -2 is outside of the range of cosine. Therefore, the only solution is \frac{4 \pi}{3}.
  8. & \qquad \qquad \quad 3 - 3 \sin^2 x = 8 \sin x  \\& \quad 3 - 3 \sin^2 x - 8 \sin x = 0 \\& \quad 3 \sin^2 x + 8 \sin x - 3 = 0 \\& (3 \sin x - 1)(\sin x + 3) = 0 \\ & 3 \sin x - 1 \ = 0 \quad \text{or} \quad \sin x+3 = 0 \\& \qquad 3 \sin x = 1 \\& \qquad \ \sin x = \frac{1}{3} \qquad \quad \sin x = -3 \\& x = 0.3398\ radians \quad \text{No solution exists} \\& x = \pi - 0.3398 = 2.8018\ radians
  9. 2 \sin x \tan x = \tan x + \sec x 2 \sin x \cdot \frac{\sin x}{\cos x} &= \frac{\sin x}{\cos x} + \frac{1}{\cos x} \\\frac{2 \sin^2 x}{\cos x} &= \frac{\sin x + 1}{\cos x} \\2 \sin^2 x &= \sin x+1 \\2 \sin^2 x - \sin x -1 &= 0 \\(2 \sin x + 1)(\sin x - 1) &= 0 \\2 \sin x + 1 &= 0 \qquad \quad \text{or} \qquad \sin x -1 = 0 \\2 \sin x &= -1 \qquad \qquad \qquad \quad \sin x = 1 \\\sin x &= - \frac{1}{2} x = \frac{7\pi}{6},\frac{11 \pi}{6} One of the solutions is not \frac{\pi}{2}, because \tan x and \sec x in the original equation are undefined for this value of x.
  10. & \qquad \ \ 2 \cos^2x + 3 \sin x - 3 = 0 \\& \ 2(1 - \sin^2x) + 3 \sin x - 3 = 0 \ \text{Pythagorean Identity} \\& \quad 2 - 2 \sin^2x + 3 \sin x - 3 = 0 \\& \quad \ -2 \sin^2x + 3 \sin x - 1 = 0\ \text{Multiply by}\ -1 \\& \qquad \ \ 2 \sin^2x - 3 \sin x + 1 = 0 \\& \quad \ \ (2 \sin x - 1)(\sin x - 1) = 0 \\& 2 \sin x - 1 \ = 0 \qquad \qquad \text{or} \qquad \sin x -1 = 0 \\& \qquad 2 \sin x = 1 \\& \qquad \ \sin x = \frac{1}{2} \qquad \qquad \qquad \qquad \ \ \sin x = 1 \\& \qquad x = \frac{\pi}{6}, \frac{5 \pi}{6} \qquad \qquad \qquad \qquad \qquad \  x = \frac{\pi}{2}
  11. \tan^2 x + \tan x - 2 = 0 \frac{-1 \pm \sqrt{1^2 - 4(1)(-2)}}{2} & = \tan x \\\frac{-1 \pm \sqrt{1 + 8}}{2} & = \tan x \\\frac{-1 \pm 3}{2} & = \tan x \\\tan x & = -2 \quad \text{or}\quad 1 \tan x = 1 when x = \frac{\pi}{4}, in the interval \left [-\frac{\pi}{2}, \frac{\pi}{2} \right ] \tan x = -2 when x = -1.107 \ rad
  12. 5 \cos^2 \theta - 6 \sin \theta = 0 over the interval [0, 2\pi]. 5 \left(1 - \sin^2 x \right ) - 6 \sin x & =0 \\-5 \sin^2 x - 6 \sin x + 5 & = 0 \\5 \sin^2 x + 6 \sin x - 5 & = 0 \\\frac{-6 \pm \sqrt{6^2 - 4(5)(-5)}}{2(5)} & = \sin x \\ \frac{-6 \pm \sqrt{36 + 100}}{10} & = \sin x \\\frac{-6 \pm \sqrt{136}}{10} & = \sin x \\\frac{-6 \pm 2 \sqrt{34}}{10} & = \sin x \\\frac{-3 \pm \sqrt{34}}{5} & = \sin x x = \sin^{-1} \left (\frac{-3 + \sqrt{34}}{5} \right ) or \sin^{-1}\left (\frac{-3 - \sqrt{34}}{5} \right ) x = 0.6018\ rad or 2.5398\ rad from the first expression, the second expression will not yield any answers because it is out the the range of sine.

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CK.MAT.ENG.SE.2.Trigonometry.3.3

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