Answer: 0.93 radians & 2.21 radians
Step-by-step explanation:
[tex]-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)[/tex]
[tex]\theta=not\ valid\qquad and\quad \theta=0.927[/tex]
[tex]\theta = 0.927\ radians\text{\ in the 1st quadrant and}\\\pi-0.927=2.21\ radians\text{\ in the 2nd quadrant}[/tex]
Answer:
2.21
0.93
Step-by-step explanation:
Given that; [tex]-4\sin^2\theta-3\sin \theta+5=0[/tex]
This is a quadratic equation is [tex]\sin \theta[/tex], where [tex]a=-4,b=-3,c=5[/tex]
We want to solve for [tex]\theta[/tex] in radians, where 0≤θ<2π.
We apply the quadratic formula given by;
[tex]\sin \theta=\frac{-b\pm\sqrt{b^2-4ac} }{2a}[/tex]
We substitute the given values to obtain;
[tex]\sin \theta=\frac{--3\pm\sqrt{(-3)^2-4(-4)(5)}}{2(-4)}[/tex]
Simplify;
[tex]\sin \theta=\frac{3\pm\sqrt{9+80}}{-8}[/tex]
[tex]\sin \theta=\frac{3\pm\sqrt{89}}{-8}[/tex]
[tex]\sin \theta=0.804[/tex] or [tex]\sin \theta=-1.55[/tex]
When [tex]\sin \theta=0.804[/tex] , [tex]\theta=\sin^{-1}(0.804)[/tex]
[tex]\Rightarrow \theta=0.93[/tex] --In the first quadrant.
In the second quadrant;
[tex]\theta=\pi-0.93=2.21[/tex]
When [tex]\sin \theta=-1.55[/tex] , [tex]\theta[/tex] is not defined.
