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When $a$ is irrational, the number of solutions satisfying the equation $1+\sin ^2 a x=\cos x$ is
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Verified Answer
The correct answer is:
1
We have,
$$
1+\sin ^2 a x=\cos x
$$
Now, $1+\sin ^2 a x \geq 1$ and $-1 \leq \cos x \leq 1$
$\therefore$ It is possible only
$$
\begin{gathered}
1+\sin ^2 a x=1 \text { at } x=2 n \pi \\
\Rightarrow \quad \sin ^2 a x=0 \Rightarrow \sin a x=0 \Rightarrow a x=n \pi
\end{gathered}
$$
$$
\Rightarrow \quad a=\frac{n \pi}{x} \Rightarrow a=\frac{n \pi}{2 n \pi} \Rightarrow a=\frac{1}{2} .
$$
But $a$ is irrational.
$\therefore$ It is possible only $x=0$.
$\because a$ is irrational only one solution satisfying the equation at $x=0$.
$$
1+\sin ^2 a x=\cos x
$$
Now, $1+\sin ^2 a x \geq 1$ and $-1 \leq \cos x \leq 1$
$\therefore$ It is possible only
$$
\begin{gathered}
1+\sin ^2 a x=1 \text { at } x=2 n \pi \\
\Rightarrow \quad \sin ^2 a x=0 \Rightarrow \sin a x=0 \Rightarrow a x=n \pi
\end{gathered}
$$
$$
\Rightarrow \quad a=\frac{n \pi}{x} \Rightarrow a=\frac{n \pi}{2 n \pi} \Rightarrow a=\frac{1}{2} .
$$
But $a$ is irrational.
$\therefore$ It is possible only $x=0$.
$\because a$ is irrational only one solution satisfying the equation at $x=0$.
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