If 1 + cosx - 2sin^2x = 0 , what is the value of x?

Jun 11, 2016

$x = \left\{\pm \pi + 2 k \pi\right\} \cup \left\{\pm \frac{\pi}{3} + 2 k \pi\right\}$ for $k = 0 , \pm 1 , \pm 2 , \pm 3 , \ldots$

Explanation:

Making $\sin {\left(x\right)}^{2} = 1 - \cos {\left(x\right)}^{2}$ and $y = \cos \left(x\right)$ we obtain

$1 + y - 2 \left(1 - {y}^{2}\right) = 0$

Solving for $y$ we have $\left\{y = - 1 , y = \frac{1}{2}\right\}$ so the solutions for $x$ must obey

$\cos \left(x\right) = - 1$ or $\cos \left(x\right) = \frac{1}{2}$

Solving for $x$ we obtain

$x = \pm \pi + 2 k \pi$ for $\cos \left(x\right) = - 1$ and
$x = \pm \frac{\pi}{3} + 2 k \pi$ for $\cos \left(x\right) = \frac{1}{2}$
for $k = 0 , \pm 1 , \pm 2 , \pm 3 , \ldots$