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Trigonometric Equations Using Factoring

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Trigonometric Equations Using Factoring

Solving trig equations is an important process in mathematics. Quite often you'll see powers of trigonometric functions and be asked to solve for the values of the variable which make the equation true. For example, suppose you were given the trig equation

2 \sin x \cos x = \cos x

Could you solve this equation? (You might be tempted to just divide both sides by \cos x , but that would be incorrect because you would lose some solutions.) Instead, you're going to have to use factoring. Read this Concept, and at its conclusion, you'll be ready to factor the above equation and solve it.

Watch This

James Sousa Example: Solve a Trig Equation by Factoring

Guidance

You have no doubt had experience with factoring. You have probably factored equations when looking for the possible values of some variable, such as "x". It might interest you to find out that you can use the same factoring method for more than just a variable that is a number. You can factor trigonometric equations to find the possible values the function can take to satisfy an equation.

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 A

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 B

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 C

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.

Vocabulary

Factoring: Factoring is a way to solve trigonometric equations by separating the equation into two terms which, when multiplied together, give the original expression. Since the product of the two factors is equal to zero, each of the factors can be equal to zero to make the original expression true. This leads to solutions for the original expression.

Guided Practice

1. Solve the trigonometric equation 4 \sin x \cos x + 2 \cos x-2 \sin x - 1 = 0 such that 0 \le x < 2\pi .

2. Solve \tan^2 x = 3 \tan x for x over [0, \pi] .

3. 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) .

Solutions:

1. 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}

2.

\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

3.

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} .

Concept Problem Solution

The equation you were given is

2 \sin x \cos x = \cos x

To solve this:

2 \sin x \cos x = \cos x

Subtract \cos x from both sides and factor it out of the equation:

2 \sin x \cos x - \cos x = 0\\\cos x (2 \sin x - 1) = 0\\

Now set each factor equal to zero and solve. The first is \cos x :

\cos x = 0\\x = \frac{\pi}{2}, \frac{3\pi}{2}\\

And now for the other term:

2 \sin x - 1 = 0\\\sin x = \frac{1}{2}\\x = \frac{\pi}{6}, \frac{5\pi}{6}\\

Practice

Solve each equation for x over the interval [0,2\pi) .

  1. \cos^2(x)+2\cos(x)+1=0
  2. 1-2\sin(x)+\sin^2(x)=0
  3. 2\cos(x)\sin(x)-\cos(x)=0
  4. \sin(x)\tan^2(x)-\sin(x)=0
  5. \sec^2(x)=4
  6. \sin^2(x)-2\sin(x)=0
  7. 3\sin(x)=2\cos^2(x)
  8. 2\sin^2(x)+3\sin(x)=2
  9. \tan(x)\sin^2(x)=\tan(x)
  10. 2\sin^2(x)+\sin(x)=1
  11. 2\cos(x)\tan(x)-\tan(x)=0
  12. \sin^2(x)+\sin(x)=2
  13. \tan(x)(2\cos^2(x)+3\cos(x)-2)=0
  14. \sin^2(x)+1=2\sin(x)
  15. 2\cos^2(x)-3\cos(x)=2

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