The number of solutions of the given equation | vedclass

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(D) The given equation is $a \sin x + b \cos x = c$.Divide both sides by $\sqrt{a^2 + b^2}$:$\frac{a}{\sqrt{a^2 + b^2}} \sin x + \frac{b}{\sqrt{a^2 +

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(D) The given equation is $a \sin x + b \cos x = c$.Divide both sides by $\sqrt{a^2 + b^2}$:$\frac{a}{\sqrt{a^2 + b^2}} \sin x + \frac{b}{\sqrt{a^2 + b^2}} \cos x = \frac{c}{\sqrt{a^2 + b^2}}$Let $\frac{a}{\sqrt{a^2 + b^2}} = \cos \alpha$ and $\frac{b}{\sqrt{a^2 + b^2}} = \sin \alpha$.Then the equation becomes $\sin(x + \alpha) = \frac{c}{\sqrt{a^2 + b^2}}$.Since $|c| > \sqrt{a^2 + b^2}$,it follows that $\left| \frac{c}{\sqrt{a^2 + b^2}} \right| > 1$.Because the range of the sine function is $[-1, 1]$,the equation $\sin(x + \alpha) = k$ has no solution if $|k| > 1$.Therefore,the given equation has no solution.

The number of solutions of the given equation $a \sin x + b \cos x = c$,where $|c| > \sqrt{a^2 + b^2}$,is

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