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Solving Trigonometric Equations Using Basic Algebra

Substitute in potential solutions or solve using inverse trig functions.

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Solving Trigonometric Equations Using Algebra

As Agent Trigonometry, you are given this clue: 2 \sin x-\sqrt{2}=0 . If 0 \le x < 2 \pi , what is/are the value(s) of x .


In the previous concept, we verified trigonometric identities, which are true for every real value of x . In this concept, we will solve trigonometric equations. An equation is only true for some values of x .

Example A

Verify that \csc x-2=0 when x=\frac{5 \pi}{6} .

Solution: Substitute in x=\frac{5 \pi}{6} to see if the equations holds true.

\csc \left(\frac{5 \pi}{6}\right)-2&=0 \\\frac{1}{\sin \left(\frac{5 \pi}{6}\right)}-2&=0 \\\frac{1}{\frac{1}{2}}-2&=0 \\2-2&=0

This is a true statement, so x=\frac{5 \pi}{6} is a solution to the equation.

Example B

Solve 2 \cos x+1=0 .

Solution: To solve this equation, we need to isolate \cos x and then use inverse to find the values of x when the equation is valid. You already did this to find the zeros in the graphing concepts earlier in this chapter.

2 \cos x+1&=0 \\2 \cos x&=-1 \\\cos x&=- \frac{1}{2}

So, when is the \cos x=- \frac{1}{2} ? Between 0 \le x< 2 \pi, x=\frac{2 \pi}{3} and \frac{4 \pi}{3} . But, the trig functions are periodic, so there are more solutions than just these two. You can write the general solutions as x=\frac{2 \pi}{3} \pm 2 \pi n and x=\frac{4 \pi}{3} \pm 2 \pi n , where n is any integer. You can check your answer graphically by graphing y=\cos x and y=- \frac{1}{2} on the same set of axes. Where the two lines intersect are the solutions.

Example C

Solve 5 \tan(x+2)-1=0 , where 0 \le x < 2 \pi .

Solution: In this example, we have an interval where we want to find x . Therefore, at the end of the problem, we will need to add or subtract \pi , the period of tangent, to find the correct solutions within our interval.

5 \tan(x+2)-1&=0 \\5 \tan(x+2)&=1 \\\tan(x+2)&=\frac{1}{5}

Using the \tan^{-1} button on your calculator, we get that \tan^{-1} \left(\frac{1}{5}\right)=0.1974 . Therefore, we have:

x+2&=0.1974 \\x&=-1.8026

This answer is not within our interval. To find the solutions in the interval, add \pi a couple of times until we have found all of the solutions in [0, 2 \pi] .

x&=-1.8026+ \pi=1.3390 \\ &=1.3390+ \pi=4.4806

The two solutions are x = 1.3390 and 4.4806.

Concept Problem Revisit To solve this equation, we need to isolate \sin x and then use inverse to find the values of x when the equation is valid.

2 \sin x -\sqrt{2}=0 \\2 \sin x&=\sqrt{2} \\\sin x&=\frac{\sqrt{2}}{2}

So now we need to find the values of x for which \sin x =\frac{\sqrt{2}}{2} . We know from the special triangles that this value of sine holds true for a 45^\circ angle, but is that the only value of x for which it is true?

We are told that 0 \le x < 2 \pi . Recall that the sine is positive in both the first and second quadrants, so \sin x =\frac{\sqrt{2}}{2} when x also is 135^\circ .

Guided Practice

1. Determine if x=\frac{\pi}{3} is a solution for 2 \sin x=\sqrt{3} .

Solve the following trig equations in the interval 0 \le x< 2 \pi .

2. 3 \cos^2x-5=0

3. 3 \sec(x-1)+2=0


1. 2 \sin \frac{\pi}{3}= \sqrt{3} \rightarrow 2 \cdot \frac{\sqrt{3}}{2}=\sqrt{3} Yes, x=\frac{\pi}{3} is a solution.

2. Isolate the \cos^2x and then take the square root of both sides. Don’t forget about the \pm !

9 \cos^2x-5&=0 \\9 \cos^2x&=5 \\\cos^2x&=\frac{5}{9} \\\cos x&=\pm \frac{\sqrt{5}}{3}

The \cos x=\frac{\sqrt{5}}{3} at x=0.243 rad (use your graphing calculator). To find the other value where cosine is positive, subtract 0.243 from 2 \pi , x=2 \pi -0.243=6.037 rad.

The \cos x=- \frac{\sqrt{5}}{3} at x=2.412 rad, which is in the 2^{nd} quadrant. To find the other value where cosine is negative (the 3^{rd} quadrant), use the reference angle, 0.243, and add it to \pi . x= \pi+0.243=3.383 rad.

3. Here, we will find the solution within the given range, 0 \le x< 2 \pi .

3 \sec(x-1)+2&=0 \\3 \sec(x-1)&=-2 \\\sec(x-1)&=- \frac{2}{3} \\\cos(x-1)&=- \frac{3}{2}

At this point, we can stop. The range of the cosine function is from 1 to -1. - \frac{3}{2} is outside of this range, so there is no solution to this equation.


Determine if the following values for x . are solutions to the equation 5+6 \csc x=17 .

  1. x=- \frac{7 \pi}{6}
  2. x=\frac{11 \pi}{6}
  3. x=\frac{5 \pi}{6}

Solve the following trigonometric equations. If no solutions exist, write no solution .

  1. 1- \cos x=0
  2. 3 \tan x - \sqrt{3}=0
  3. 4 \cos x=2 \cos x+1
  4. 5 \sin x-2=2 \sin x+4
  5. \sec x-4=- \sec x
  6. \tan^2(x-2)=3

Sole the following trigonometric equations within the interval 0 \le x < 2 \pi . If no solutions exist, write no solution .

  1. \cos x=\sin x
  2. - \sqrt{3} \csc x=2
  3. 6 \sin(x-2)=14
  4. 7 \cos x -4=1
  5. 5+4 \cot^2x=17
  6. 2 \sin^2x-7=-6

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