When a is irrational,the number of solutions | vedclass

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(A) We have the equation $1+\sin^2(ax)=\cos(x)$.Since $\sin^2(ax) \geq 0$,we have $1+\sin^2(ax) \geq 1$.Also,we know that $\cos(x) \leq 1$ for all $x

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(A) We have the equation $1+\sin^2(ax)=\cos(x)$.Since $\sin^2(ax) \geq 0$,we have $1+\sin^2(ax) \geq 1$.Also,we know that $\cos(x) \leq 1$ for all $x \in \mathbb{R}$.For the equality $1+\sin^2(ax)=\cos(x)$ to hold,both sides must be equal to $1$.Thus,we must have $\cos(x) = 1$,which implies $x = 2n\pi$ for some integer $n$.Substituting $x = 2n\pi$ into the equation,we get $1+\sin^2(a(2n\pi)) = 1$,which implies $\sin^2(2an\pi) = 0$.This means $\sin(2an\pi) = 0$,so $2an\pi = k\pi$ for some integer $k$,which simplifies to $2an = k$,or $a = \frac{k}{2n}$.If $n 
eq 0$,then $a$ must be a rational number.However,the problem states that $a$ is irrational.Therefore,the only possible value for $n$ is $0$,which gives $x = 2(0)\pi = 0$.Checking $x=0$: $1+\sin^2(a \cdot 0) = 1+\sin^2(0) = 1+0 = 1$,and $\cos(0) = 1$.Thus,$x=0$ is the only solution.

When $a$ is irrational,the number of solutions satisfying the equation $1+\sin^2(ax)=\cos(x)$ is

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