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 $\displaystyle \sin 2x = 2 \sin x \cos x$ to replace $\displaystyle \sin 2x$
Then,
$\displaystyle 2 \sin x \cos x - \sin x = 2 \cos^2 x - \cos x$
$\displaystyle 2 \sin x \cos x - \sin x - 2 \cos^2 x + \cos x = 0$
$\displaystyle \sin x (2 \cos x - 1) - \cos x(2 \cos x - 1)=0$
$\displaystyle (\sin x - \cos x)(2 \cos x - 1)=0$
$\displaystyle \sin x - \cos x = 0 \text{ or } 2 \cos x -1 = 0$
In which case,
$\displaystyle \sin x = \cos x \text{ or } \cos x = \frac{1}{2}$
In the first case, $\displaystyle \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, $\displaystyle \cos x = \frac{1}{2} \text{ at } \boxed{\frac{\pi}{3}+2\pi k \text{ and at }\frac{5\pi}{3}+2\pi k}$
My answers differ from Skeeter's. Who is correct?