Trigonometric equations

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If $x \neq \frac{n π}{2}$ and ${(\cos x)}^{\sin^{2} x - 3 \sin x + 2} = 1$ then the general solutions of x is

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Solution

As $x \neq \frac{n π}{2}$ $\Rightarrow \cos x \neq 0, 1 - 1$
So, ${(\cos x)}^{\sin^{2} x - 3 \sin x + 2} = 1$
$\Rightarrow \sin^{2} x - 3 \sin x + 2 = 0$
$\Rightarrow (\sin x - 2) (\sin x - 1) = 0$
$\Rightarrow \sin x = 1, 2$
Where $\sin x = 2$ is not possible, and $\sin x = 1 \Rightarrow x = 2 n π + \frac{π}{2}$ does not satisfy the given condition.

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Trigonometric equations

Remember concepts with our Masterclasses.

If x≠nπ2and cosxsin2x−3sinx+2=1 then the general solutions of x is

As x≠nπ2⇒cosx≠0,1−1So, cosxsin2x−3sinx+2=1⇒sin2x−3sinx+2=0⇒sinx−2sinx−1=0⇒sinx=1,2Where sinx=2 is not possible, and sinx=1⇒x=2nπ+π2 does not satisfy the given condition.

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If $x \neq \frac{n π}{2}$ and ${(\cos x)}^{\sin^{2} x - 3 \sin x + 2} = 1$ then the general solutions of x is