# Solving Trigonometric Equations.

#### Cristina999

Solve the trigonometric equation. For each find the exact solutions in [0,2pi]

A. Sin(3x)=-1/2
B.COS (2x)=COS (x)

#### chiro

MHF Helper
Hey Cristina999.

What have you tried?

#### HallsofIvy

MHF Helper
"Sine", if I recall correctly, is "opposite side over hypotenuse" so I might start by drawing a right triangle with leg opposite the angle equal to half the hypotenuse. Then I might realize that If I put two of those together, with the third sides as common side, I have a triangle with all three sides the same length- two sides are the two hypotenuses and the third side is made with the two 1/2 sides. What can you say about the angles in a equilateral triangle?

For the other problem, what do you know about when two angles have the same cosine? Think about the graph of y= cos(x).

#### skeeter

MHF Helper
from your intimate knowlege of the unit circle, you should already know that sine has a value of $-\dfrac{1}{2}$ at the angles $\dfrac{7\pi}{6}$ in quad III and $\dfrac{11\pi}{6}$ in quad IV.

$0 \le x < 2\pi \implies 0 \le 3x < 6\pi$

$3x = \bigg[\dfrac{7\pi}{6} \, , \, \dfrac{11\pi}{6} \, , \, \dfrac{19\pi}{6} \, , \, \dfrac{23\pi}{6} \, , \, \dfrac{31\pi}{6}\, , \, \dfrac{35\pi}{6} \bigg]$

finish solving for $x$.

For the second problem, note the double angle identity $\cos(2x)=2\cos^2{x}-1$ ...

$2\cos^2{x}-1=\cos{x}$

$2\cos^2{x}-\cos{x}-1=0$

$(2\cos{x}+1)(\cos{x}-1)=0$

use the zero product property ...