Question
If $2\sin^2\theta-\cos^2\theta=2,$ then find the value of $\theta$.
✓
Answer
Given, $2\sin^2\theta-\cos^2\theta=2$
$\Rightarrow\ 2\sin^2\theta-(1-\sin^2\theta)=2 \ \big[\because\ \sin^2\theta+\cos^2\theta=1\big]$
$\Rightarrow\ \sin^2\theta+\sin^2\theta-1=2$
$\Rightarrow\ 3\sin^2\theta=3$
$\Rightarrow\ \sin^2\theta=1\ \big[\because\ \sin90^\circ=1\big]$
$\Rightarrow\ \sin\theta=1=\sin90^\circ$
$\therefore\ \theta=90^\circ$
$\Rightarrow\ 2\sin^2\theta-(1-\sin^2\theta)=2 \ \big[\because\ \sin^2\theta+\cos^2\theta=1\big]$
$\Rightarrow\ \sin^2\theta+\sin^2\theta-1=2$
$\Rightarrow\ 3\sin^2\theta=3$
$\Rightarrow\ \sin^2\theta=1\ \big[\because\ \sin90^\circ=1\big]$
$\Rightarrow\ \sin\theta=1=\sin90^\circ$
$\therefore\ \theta=90^\circ$
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