Question

i have an advanced trig equation problem i need help with. pic included

Answer 1

In general,

$\begin{gathered} \cos (2x)=\cos (x+x)=\cos x\cos x-\sin x\sin x=\cos ^2x-\sin ^2x \\ \Rightarrow\cos (2x)=\cos ^2x-\sin ^2x \end{gathered}$

Therefore, the initial equation becomes

$\begin{gathered} \Rightarrow9(\cos ^2x-\sin ^2x)=9\sin ^2x+5 \\ \Rightarrow9\cos ^2x=18\sin ^2x+5 \end{gathered}$

Furthermore,

$\begin{gathered} \sin ^2x+\cos ^2x=1 \\ \Rightarrow\cos ^2x=1-\sin ^2x \end{gathered}$

Thus,

$\begin{gathered} \Rightarrow9(1-\sin ^2x)=18\sin ^2x+5 \\ \Rightarrow9-9\sin ^2x=18\sin ^2x+5 \\ \Rightarrow27\sin ^2x=9-5=4 \\ \Rightarrow\sin ^2x=\frac{4}{27} \end{gathered}$$\Rightarrow\sin x=\pm\frac{2}{3\sqrt[]{3}}$

Finally,

$\begin{gathered} \Rightarrow x=\pi n\pm\sin ^{-1}(\frac{2}{3\sqrt[]{3}}) \\ \Rightarrow x=\pi n+0.39510,n\in Z \\ \text{and} \\ x=\pi n-0.39510 \end{gathered}$

Finding the values of x on [0,2pi)

$\Rightarrow x=0.39510,\pi-0.39510,\pi+0.39510,2\pi-0.39510$

Rounding to 3 decimal places,

$\Rightarrow x=0.395,2.746,3.537,5.888$

The four answers are shown above