# Thread: [Pre-Calc] Finding all solutions of a trig equation

1. ## [Pre-Calc] Finding all solutions of a trig equation

sin(2x) - sinx = 2cos^2 (x) - cosx

Can someone show me the steps on how to find all the solutions in the interval [0,2pi)?

2. ## Re: [Pre-Calc] Finding all solutions of a trig equation

Originally Posted by albo23
sin(2x) - sinx = 2cos^2 (x) - cosx

Can someone show me the steps on how to find all the solutions in the interval [0,2pi)?
$2\sin{x}\cos{x} - \sin{x} = 2\cos^2{x} - \cos{x}$

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

edit ... pulled out the factor incorrectly.

\$#it happens.

3. ## Re: [Pre-Calc] Finding all solutions of a trig equation

Originally Posted by albo23
sin(2x) - sinx = 2cos^2 (x) - cosx

Can someone show me the steps on how to find all the solutions in the interval [0,2pi)?
Hi albo23,

Slightly different approach. Use the double angle identity $\sin 2x = 2 \sin x \cos x$ to replace $\sin 2x$

Then,

$2 \sin x \cos x - \sin x = 2 \cos^2 x - \cos x$

$2 \sin x \cos x - \sin x - 2 \cos^2 x + \cos x = 0$

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

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

$\sin x - \cos x = 0 \text{ or } 2 \cos x -1 = 0$

In which case,

$\sin x = \cos x \text{ or } \cos x = \frac{1}{2}$

In the first case, $\sin x = \cos x \text{ at } \boxed{\frac{\pi}{4}+2\pi k \text{ and at }\frac{5\pi}{4}+2\pi k}$

In the second case, $\cos x = \frac{1}{2} \text{ at } \boxed{\frac{\pi}{3}+2\pi k \text{ and at }\frac{5\pi}{3}+2\pi k}$

My answers differ from Skeeter's. Who is correct?

4. ## Re: [Pre-Calc] Finding all solutions of a trig equation

Originally Posted by masters
My answers differ from Skeeter's. Who is correct?
Skeeter got a bit carried away.
$\color{red}{\cos{x}(2\sin{x} - 1)} - \cos{x}(2\cos{x} - 1) = 0$