Question #d574e

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Answer 1 · 2016-05-05T05:22:32

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

Solve the trigonometric equation

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

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

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

Square both sides of the equation

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

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

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

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

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

Quadratic Equation

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

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

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

then solve for w

$w = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $w=(-b+-sqrt(b^2-4ac))/(2a)$

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

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

$w_{1} = \frac{7 + 3}{20} = \frac{1}{2}$ $w_1=(7+3)/20=1/2$
$w_{2} = \frac{7 - 3}{20} = \frac{1}{5}$ $w_2=(7-3)/20=1/5$

${sin}^{2} x = \pm \sqrt{\frac{1}{2}}$ $sin^2 x=+-sqrt(1/2)$ and

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

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

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