# How to find the trigonometric equation for 3(cos x - 2sin^2 x) = cos x - 2 ?

Aug 7, 2018

$2 \cos x - 3 \cos 2 x = 1$

#### Explanation:

$3 \left(\cos x - 2 {\sin}^{2} x\right) = \cos x - 2$

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

$\therefore 3 \left(\cos x - \cos 2 x - 1\right) = \cos x - 2$

$3 \cos x - 3 \cos 2 x - 3 = \cos x - 2$

$3 \cos x - \cos x - 3 \cos 2 x = 3 - 2$

$2 \cos x - 3 \cos 2 x = 1$

Aug 7, 2018

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

#### Explanation:

Using ${\sin}^{2} x = 1 - {\cos}^{2} x \mathmr{and} c = \cos x$,

$6 {c}^{2} + 2 c - 4 = 2 \left(3 {c}^{2} + c - 2\right) = 0$

#rArr (cos x+1)( cos x - 2/3 ) = 0