# Question #d574e

$x = {45}^{\circ}$is the correct value of $x$
If at $x = {26.56505118}^{\circ}$ then $\tan x = \frac{1}{2}$

#### Explanation:

Solve the trigonometric equation

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

Let $\cos x = \sqrt{1 - {\sin}^{2} x} \text{ }$from Pythagorean Relation

$1 + {\sin}^{2} x = 3 \sin x \sqrt{1 - {\sin}^{2} x}$

Square both sides of the equation

$1 + {\sin}^{2} x = 3 \sin x \sqrt{1 - {\sin}^{2} x}$

${\left(1 + {\sin}^{2} x\right)}^{2} = {\left(3 \sin x \sqrt{1 - {\sin}^{2} x}\right)}^{2}$

$1 + 2 \cdot {\sin}^{2} x + {\sin}^{4} x = 9 \cdot {\sin}^{2} x \left(1 - {\sin}^{2} x\right)$

$1 + 2 \cdot {\sin}^{2} x + {\sin}^{4} x = 9 \cdot {\sin}^{2} x - 9 \cdot {\sin}^{4} x$

$10 \cdot {\sin}^{4} x - 7 \cdot {\sin}^{2} x + 1 = 0$

Let ${w}^{2} = {\sin}^{4} x$ and $w = {\sin}^{2} x$ then we have

$10 \cdot {\sin}^{4} x - 7 \cdot {\sin}^{2} x + 1 = 0$

$10 \cdot {w}^{2} - 7 \cdot w + 1 = 0$

then solve for w

$w = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$w = \frac{- \left(- 7\right) \pm \sqrt{{\left(- 7\right)}^{2} - 4 \left(10\right) \left(1\right)}}{2 \left(10\right)}$

$w = \frac{7 \pm \sqrt{49 - 40}}{2 \left(10\right)}$

${w}_{1} = \frac{7 + 3}{20} = \frac{1}{2}$
${w}_{2} = \frac{7 - 3}{20} = \frac{1}{5}$

${\sin}^{2} x = \pm \sqrt{\frac{1}{2}}$ and

$x = {45}^{\circ}$ which is the root of the equation

$x = - {45}^{\circ}$ not a root
$x = \pm {26.565}^{\circ}$ not a root

God bless...I hope the explanation is useful.