# Find the exact value? 3-2sin^2x=3cosx

May 9, 2018

$\rightarrow x = 2 n \pi \pm \frac{\pi}{3}$ OR $x = 2 n \pi$

#### Explanation:

$\rightarrow 3 - 2 {\sin}^{2} x = 3 \cos x$

$\rightarrow 3 - 2 \left(1 - {\cos}^{2} x\right) = 3 \cos x$

$\rightarrow 3 - 2 + 2 {\cos}^{2} x = 3 \cos x$

$\rightarrow 2 {\cos}^{2} x - 3 \cos x + 1 = 0$

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

$\rightarrow 2 \cos x \left(\cos x - 1\right) - 1 \left(\cos x - 1\right) = 0$

$\rightarrow \left(2 \cos x - 1\right) \left(\cos x - 1\right) = 0$

Either, $2 \cos x - 1 = 0$

$\rightarrow \cos x = \frac{1}{2} = \cos \left(\frac{\pi}{3}\right)$

$\rightarrow x = 2 n \pi \pm \frac{\pi}{3}$ where $n \rightarrow Z$

OR, $\cos x - 1 = 0$

$\rightarrow \cos x = 1 = \cos 0$

$\rightarrow x = 2 \pi n$ where $n \rightarrow Z$

Jun 2, 2018

$x = \pm {60}^{\circ} + {360}^{\circ} k \mathmr{and} x = {360}^{\circ} k \quad$ integer $k$

#### Explanation:

The other answer is fine; I just want to add this to my collection of yet anther trig problem that's going to work out to 30/60/90 or 45/45/90. When are we going to realize that it's goofy to teach a subject that only handles two triangles reasonably well?

$3 - 2 {\sin}^{2} x = 3 \cos x$

$3 - 2 \left(1 - {\cos}^{2} x\right) = 3 \cos x$

$2 {\cos}^{2} x - 3 \cos x + 1 = 0$

$\left(2 \cos x - 1\right) \left(\cos x - 1\right) = 0$

$\cos x = \frac{1}{2} \mathmr{and} \cos x = 1$

Cosine or sine of $\frac{1}{2}$; same old same old.

$\cos x = \cos {60}^{\circ} \mathmr{and} \cos x = \cos 0$

$x = \pm {60}^{\circ} + {360}^{\circ} k \mathmr{and} x = {360}^{\circ} k \quad$ integer $k$