Question

Given sine of theta equals negative 7 over 25 and pi is less than theta is less than 3 times pi over 2 comma what is the exact solution of sin 2θ?

Answer 1

$\begin{gathered} \sin (\theta)=-\frac{7}{25} \\ \pi<\theta<\frac{3\pi}{2} \end{gathered}$

Therefore, according to this, the angle is in the III quadrant, so:

$\sin (2\theta)=2\sin (\theta)\cos (\theta)$

Where:

$\cos (\theta)=\frac{\sqrt[]{25^2-7^2}}{25}=-\frac{24}{25}$

Replace the values:

$\begin{gathered} \sin (2\theta)=2(-\frac{7}{25})\cdot(-\frac{24}{25}) \\ \sin (2\theta)=\frac{336}{625}=0.5376 \end{gathered}$$\begin{gathered} f(x)=\cos (2x)+1 \\ g(x)=\sin (x)+2 \end{gathered}$

In order to find where the functions intersect we need to equal the functions:

$\cos (2x)+1=\sin (x)+2$

Replace cos(2x) = 1 - 2sin²(x) using the identity of the double angle:

$\begin{gathered} 1-2\sin ^2(x)+1=\sin (x)+2 \\ so\colon \\ 2\sin ^2(x)+\sin (x)=0 \\ \sin (x)(1+2\sin (x))=0 \end{gathered}$

Split into two equations:

$\begin{gathered} \sin (x)=0 \\ x=\pi\cdot n1_{\text{ }},_{\text{ }}n1\in\Z \\ ---------- \\ 2\sin (x)+1=0 \\ \sin (x)=-\frac{1}{2} \\ x=2\pi n2+\frac{7\pi}{6};_{\text{ }}n2\in\Z \\ x=2\pi n3+\frac{11\pi}{6};_{\text{ }}n2\in\Z \end{gathered}$

Since the solutions are in the interval [0, 2π):

$x=0,\pi,\frac{7\pi}{6},\frac{11\pi}{6}$

Here is a graph where you can verify the answer.