# Solving Trigonometric Equations

Definition: To solve trigonometric equations, we’ll need to combine our knowledge of the unit circle, trigonometric functions, and trigonometric identities with algebraic methods of solving.

Example 1: Find all solutions of cos(2x) = 0 in the interval [0, 2π).

Solution:

Step 1: Recall where the cosine function has a value of 0. We know this occurs on the y-axis of the unit circle at π/2, 3π/2, and for any value π/2 + nπ (where n is an integer).

Looking at the first value, π/2,  we can write the following equation and solve for x:

$\dpi{120} \bg_white \\2x = \frac{\pi}{2}\\\\ x = \frac{1}{2}\left (\frac{\pi}{2}\right )\\\\ x = \frac{\pi}{4}$

Step 2: Repeat this step to find all solutions in the given interval, [0, 2π), keeping in mind the general solution π/2 + nπ :

$\dpi{120} \bg_white \\2x = \frac{3\pi}{2}\\\\ {\;\;}x= \frac{3\pi}{4}\\\\\\2x = \frac{5\pi}{2}\\\\ {\;\;}x= \frac{5\pi}{4}\\\\\\2x = \frac{7\pi}{2}\\\\ {\;\;}x= \frac{7\pi}{4}$

This gives us the following solutions:

$\dpi{120} \bg_white x =\;\; \frac{\pi}{4},\; \frac{3\pi}{4},\; \frac{5\pi}{4},\; \frac{7\pi}{4}$

Example 2: Find all solutions:

$\dpi{120} \bg_white 16sin^{4}\,x - 16sin^2\,x + 3 = 0$

Solution:

Step 1: We will need to solve the sine function as if it were a quadratic, so can start  by factoring and applying the zero product principle:

$\dpi{120} \bg_white \\16sin^{4}\,x - 16sin^2\,x + 3 = 0\\\\ (4sin^2\,x - 3)(4sin^2\,x -1) = 0\\\\ (4sin^2\,x - 3) = 0{\;\;\;\;\;\;\;\;\;\;\;\;}(4sin^2\,x - 1) = 0\\\\ {\;\;\;\;\;\;\;\;\;\;\;}sin^2\,x = \frac{3}{4}{\;\;\;\;\;\;\;\;\;\,\;\;\;\;\;\;\;\;\;\;\;\;}sin^2\,x = \frac{1}{4}\\\\ {\;\;\;\;\;\;\;\;\;\;\;\;}sin\,x = \pm \frac{\sqrt3}{2}{\;\;\;\,\,\,\;\;\;\;\;\;\;\;\;\;\;\;\;}sin\,x = \pm \frac{1}{2}$

Step 2: Solve both equations for x:

sin(x) = ±√3/2 when x = π/3, 2π/3, 4π/3, 5π/3, and all values given by the general solutions π/3 + nπ and 2π/3 + nπ (where n is an integer).

sin(x) = ±1/2 when x = π/6, 5π/6, 7π/6, 11π/6, and all values given by the general solutions  π/6 + nπ and 5π/6 + nπ, (where n is an integer).

We add multiples of π because every 2π, it’s the same angle and every odd π gives the additive inverse of the value.

Step 3: Express all of the possible solutions found in Step 2 as the final solution:

$\dpi{120} \bg_white x =\;\; \frac{\pi}{3} + n\pi,\; \frac{2\pi}{3} + n\pi,\; \frac{\pi}{6} + n\pi,\; or\; \frac{5\pi}{6} + n\pi$, (where n is an integer).

Trigonometry can be tricky! Watch and see how a Yup tutor shows a student like you how to find the general (and principle) solutions for 3 tan3(x) = tan(x) in the 0 to 2pi interval.

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