Question

URGENT!! Solve the following for θ, in radians, where 0≤θ<2π.

Answer 1

Answer:   0.93 radians & 2.21 radians

Step-by-step explanation:

$-4sin^2\theta-3sin\theta+5=0\\\\\text{Since this is not factorable, use the quadratic formula to find the roots:}\\\\sin\theta=\dfrac{-(-3)\pm \sqrt{(-3)^2-4(-4)(5)}}{2(-4)}\\\\\\.\quad=\dfrac{3\pm \sqrt{9+80}}{-8}\\\\\\.\quad=\dfrac{3\pm\sqrt{89}}{-8}\\\\\\.\quad=\dfrac{3\pm9.43}{-8}\\\\\\.\quad=\dfrac{12.43}{-8}\quad and\quad \dfrac{-6.43}{-8}\\\\\\.\quad=-1.55\quad and\quad 0.80\\\\\\\theta=sin^{-1}(-1.55)\quad and\quad \theta=sin^{-1}(0.80)$

$\theta=not\ valid\qquad and\quad \theta=0.927$

$\theta = 0.927\ radians\text{\ in the 1st quadrant and}\\\pi-0.927=2.21\ radians\text{\ in the 2nd quadrant}$

Answer 2

Answer:

2.21

0.93

Step-by-step explanation:

Given that; $-4\sin^2\theta-3\sin \theta+5=0$

This is a quadratic equation is $\sin \theta$, where $a=-4,b=-3,c=5$

We want to solve for $\theta$ in radians, where 0≤θ<2π.

We apply the quadratic formula given by;

$\sin \theta=\frac{-b\pm\sqrt{b^2-4ac} }{2a}$

We substitute the given values to obtain;

$\sin \theta=\frac{--3\pm\sqrt{(-3)^2-4(-4)(5)}}{2(-4)}$

Simplify;

$\sin \theta=\frac{3\pm\sqrt{9+80}}{-8}$

$\sin \theta=\frac{3\pm\sqrt{89}}{-8}$

$\sin \theta=0.804$ or $\sin \theta=-1.55$

When $\sin \theta=0.804$ , $\theta=\sin^{-1}(0.804)$

$\Rightarrow \theta=0.93$ --In the first quadrant.

In the second quadrant;

$\theta=\pi-0.93=2.21$

When $\sin \theta=-1.55$ , $\theta$ is not defined.