SOLUTION:
Case: Trigonometry
Given:
[tex]\begin{gathered} f(x)=2\sin^2x-1 \\ g(x)=cosx \end{gathered}[/tex]Required: To find the region when the heights are the same
Method:
Step 1: First we equate both functions
[tex]2\sin^2x-1=cosx[/tex]Step 2: Remember
[tex]\sin^2x=1-\cos^2x[/tex]Replace in the major equation
Step 3:
[tex]\begin{gathered} 2\sin^2x-1=cosx \\ 2(1-\cos^2x)-1=cosx \\ 2-2\cos^2x-1=cosx \\ Substitute\text{ p for }\cos x \\ 2-2p^2-1=p \\ 2p^2+p+1-2=0 \\ 2p^2+p-1=0 \\ Factorizing \\ 2p^2+2p-p-1=0 \\ 2p(p^+1)-1(p+1)=0 \\ (2p-1)(p+1)=0 \\ 2p-1=0\text{ or p+1=0} \\ 2p=1\text{ or p=-1} \\ p=\frac{1}{2}\text{ or p=-1} \end{gathered}[/tex]Substituting p=1/2 for cosx=p
[tex]\begin{gathered} \cos x=\frac{1}{2} \\ x=\cos^{-1}(\frac{1}{2}) \\ x=\frac{\pi}{3} \\ Cosine\text{ is also postive on the 4th quadrant} \\ (2\pi-\frac{\pi}{3}) \\ \Rightarrow\frac{5\pi}{3} \end{gathered}[/tex]Also Substituting p=-1 for cosx=p
[tex]\begin{gathered} \cos x=-1 \\ x=\cos^{-1}(-1) \\ x=\pi \end{gathered}[/tex]Final answer: Option (B)
The answers are:
[tex](\frac{\pi}{3},\pi,\frac{5\pi}{3})[/tex]