find the values of p for which the eq sinx+pcosx=2p has a solution

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Blazing goIITian

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Let sin x + pcos x = rsin(x+alpha)=rsin xcosalpha + rcos x sin alpha

So rcosalpha = 1 and rsinalpha = p

Suqring both and adding we get,

$r^2(cos^2alpha + sin^2alpha) = 1+p^2Rightarrow r = sqrt{1+p^2}$

So,

$sqrt{1+p^2}sin(x+alpha) = 2pRightarrow sin(x+alpha) = rac{2p}{sqrt{1+p^2}}$

This has a solution if $-1 le rac{2p}{sqrt{1+p^2}} le 1$

$rac{2p}{sqrt{1+p^2}} leq 1 Rightarrow 4p^2 leq 1+p^2 Rightarrow p leq rac{1}{sqrt 3}$

If p is negative,

$-p \leq\frac1{\sqrt 3} \Rightarrow p \geq -\frac{1}{\sqrt 3}$

So we have,

-rac{1}{sqrt 3} leq p leq rac{1}{sqrt 3}

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