# Solving Trigonometric Equation. Rewrite the equation 3cscx - sinx = 2 in terms of sine. Then, solve algebraically for 0 < x < 2pi?

## Sorry for asking too many questions.

Apr 2, 2018

${x}_{{1}_{1}} = \frac{\pi}{2} \mathmr{and} {x}_{{1}_{2}} = - \frac{7}{4} \pi$

#### Explanation:

$3 \csc x - \sin x = 2 | \cdot \sin \left(x\right)$
$3 - \sin {\left(x\right)}^{2} = 2 \sin \left(x\right) | + \sin {\left(x\right)}^{2} | - 3$
$0 = \sin {\left(x\right)}^{2} + 2 \sin \left(x\right) + 1 - 1 - 3$
$0 = {\left(\sin \left(x\right) + 1\right)}^{2} - 4 | + 4$
$4 = {\left(\sin \left(x\right) + 1\right)}^{2} | \sqrt{}$
$\pm 2 = \sin \left(x\right) + 1 | - 1$
$- 1 \pm 2 = \sin \left(x\right)$

$\arcsin \left(- 1 \pm 2\right) = x$
x_1=arcsin(1)+2pin, n∈ZZ
x_1=pi/2+2pin, n∈ZZ

${x}_{2} = \arcsin \left(- 3\right) + 2 \pi n$
$\arcsin \left(- 3\right) \text{ can't be expressed with real numbers.}$

arcsin(-3) and x_2 ∈CC

$\text{For } 0 < x < 2 \pi :$
${x}_{{1}_{1}} = \frac{\pi}{2} + 2 \pi \cdot 0 = \frac{\pi}{2} \mathmr{and} {x}_{{1}_{2}} = \frac{\pi}{2} - 1 \cdot 2 \pi = - \frac{7}{4} \pi$

Apr 3, 2018

$\frac{\pi}{2}$

#### Explanation:

3csc x - sin x = 2
$\frac{3}{\sin x} - \sin x = 2$
$3 - {\sin}^{2} x = 2 \sin x$
Solve this quadratic equation for sin x
${\sin}^{2} x + 2 \sin x - 3 = 0$
Since a + b + c = 0, use shortcut. There are 2 real roots:
sin x = 1 , and $\sin x = \frac{c}{a} = - 3$ (rejected)
sin x = 1 --> $x = \frac{\pi}{2}$