# How do you solve 20sinxcosxsecx - 3cos^2x= 3sin^2x?

Sep 26, 2016

$x = {\sin}^{- 1} \left(0.15\right) = {8.627}^{o}$, nearly, is the principal value.

The general value $x = {\left(n \left(180\right) + {\left(- 1\right)}^{n} 8.627\right)}^{o} , n = 0 , \pm 1 , \pm 2 , \pm 3. \ldots$

#### Explanation:

Use $\sin 2 x = 2 \sin x \mathmr{and} \cos x \mathmr{and} {\sin}^{2} x + {\cos}^{2} x = 1$.

Here, the given equation simplifies to

$\sin x = \frac{3}{20} = 0.15$. So,

$x = {\sin}^{- 1} \left(0.15\right) = {8.627}^{o}$, nearly, is the principal value.

The general value $x = {\left(n \left(180\right) + {\left(- 1\right)}^{n} 8.627\right)}^{o} , n = 0 , \pm 1 , \pm 2 , \pm 3. \ldots$

Sep 26, 2016

$\frac{10 \times 2 \sin x \cos x}{\cos} x - 3 {\cos}^{2} x = 3 {\sin}^{2} x$

$\frac{20 \sin x \cos x}{\cos} x - 3 {\cos}^{2} x = 3 {\sin}^{2} x$

$20 \sin x - 3 {\cos}^{2} x = 3 {\sin}^{2} x$

$20 \sin x = 3 {\sin}^{2} x + 3 {\cos}^{2} x$

$20 \sin x = 3 \left({\sin}^{2} x + {\cos}^{2} x\right)$

$20 \sin x = 3$

$\sin x = \frac{3}{20}$

x~= 8.6˚ and 171.4˚

Hopefully this helps!