Answer

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Hint: Use simple trigonometric formulas.

As we know that,

$

\cos 2x = 1 - 2{\sin ^2}x \\

\therefore {\sin ^2}x = \dfrac{{1 - \cos 2x}}{2} \\

$

Given: $\sin x + \dfrac{1}{2}\cos x = {\sin ^2}\left( {x + \dfrac{\pi }{4}} \right)$

Substituting the value of \[\sin x\], we get

$

\Rightarrow \sin x + \dfrac{1}{2}\cos x = \dfrac{1}{2}\left( {1 - \cos \left( {2x + \dfrac{\pi }{2}} \right)} \right) \\

\Rightarrow \sin x + \dfrac{1}{2}\cos x = \dfrac{1}{2}\left( {1 + \sin 2x} \right){\text{ }}\left( {\because \cos \left( {\dfrac{\pi }{2} + \theta } \right) = - \sin \theta } \right) \\

\Rightarrow \dfrac{{2\sin x + \cos x}}{2} = \dfrac{{1 + \sin 2x}}{2} \\

\Rightarrow 2\sin x + \cos x = 1 + \sin 2x \\

$

Now, putting $\sin 2x = 2\sin x\cos x$ in above equation, we get

$

\Rightarrow 2\sin x + \cos x = 1 + 2\sin x\cos x \\

\Rightarrow 2\sin x\cos x - 2\sin x - \cos x + 1 = 0 \\

$

Now taking $2\sin x$ common, we get

$

\Rightarrow 2\sin x\left( {\cos x - 1} \right) - 1\left( {\cos x - 1} \right) = 0 \\

\Rightarrow \left( {2\sin x - 1} \right)\left( {\cos x - 1} \right) = 0 \\

$

Either $\left( {2\sin x - 1} \right) = 0$ or $\left( {\cos x - 1} \right) = 0$.

If we take $\left( {2\sin x - 1} \right) = 0$, then

$

\Rightarrow \left( {2\sin x - 1} \right) = 0 \\

\Rightarrow 2\sin x = 1 \\

\Rightarrow \sin x = \dfrac{1}{2} \\

\Rightarrow x = \dfrac{\pi }{6},\dfrac{{5\pi }}{6} \\

$

And if we take $\left( {\cos x - 1} \right) = 0$, then

$

\Rightarrow \left( {\cos x - 1} \right) = 0 \\

\Rightarrow \cos x = 1 \\

\Rightarrow x = {0^ \circ } \\

$

Now, $x \in \left[ {0,\pi } \right]$is given.

So, the sum of all $x$ is.

\[

x = \dfrac{\pi }{6} + \dfrac{{5\pi }}{6} + 0 \\

x = \dfrac{{6\pi }}{6} \\

x = \pi \\

\]

Hence, the correct option is C.

Note: Whenever there are trigonometric equations with range given for the angle, always try to solve them by using trigonometric formulas and identities. Also, try to remember the values of basic trigonometric functions with simple angles.

As we know that,

$

\cos 2x = 1 - 2{\sin ^2}x \\

\therefore {\sin ^2}x = \dfrac{{1 - \cos 2x}}{2} \\

$

Given: $\sin x + \dfrac{1}{2}\cos x = {\sin ^2}\left( {x + \dfrac{\pi }{4}} \right)$

Substituting the value of \[\sin x\], we get

$

\Rightarrow \sin x + \dfrac{1}{2}\cos x = \dfrac{1}{2}\left( {1 - \cos \left( {2x + \dfrac{\pi }{2}} \right)} \right) \\

\Rightarrow \sin x + \dfrac{1}{2}\cos x = \dfrac{1}{2}\left( {1 + \sin 2x} \right){\text{ }}\left( {\because \cos \left( {\dfrac{\pi }{2} + \theta } \right) = - \sin \theta } \right) \\

\Rightarrow \dfrac{{2\sin x + \cos x}}{2} = \dfrac{{1 + \sin 2x}}{2} \\

\Rightarrow 2\sin x + \cos x = 1 + \sin 2x \\

$

Now, putting $\sin 2x = 2\sin x\cos x$ in above equation, we get

$

\Rightarrow 2\sin x + \cos x = 1 + 2\sin x\cos x \\

\Rightarrow 2\sin x\cos x - 2\sin x - \cos x + 1 = 0 \\

$

Now taking $2\sin x$ common, we get

$

\Rightarrow 2\sin x\left( {\cos x - 1} \right) - 1\left( {\cos x - 1} \right) = 0 \\

\Rightarrow \left( {2\sin x - 1} \right)\left( {\cos x - 1} \right) = 0 \\

$

Either $\left( {2\sin x - 1} \right) = 0$ or $\left( {\cos x - 1} \right) = 0$.

If we take $\left( {2\sin x - 1} \right) = 0$, then

$

\Rightarrow \left( {2\sin x - 1} \right) = 0 \\

\Rightarrow 2\sin x = 1 \\

\Rightarrow \sin x = \dfrac{1}{2} \\

\Rightarrow x = \dfrac{\pi }{6},\dfrac{{5\pi }}{6} \\

$

And if we take $\left( {\cos x - 1} \right) = 0$, then

$

\Rightarrow \left( {\cos x - 1} \right) = 0 \\

\Rightarrow \cos x = 1 \\

\Rightarrow x = {0^ \circ } \\

$

Now, $x \in \left[ {0,\pi } \right]$is given.

So, the sum of all $x$ is.

\[

x = \dfrac{\pi }{6} + \dfrac{{5\pi }}{6} + 0 \\

x = \dfrac{{6\pi }}{6} \\

x = \pi \\

\]

Hence, the correct option is C.

Note: Whenever there are trigonometric equations with range given for the angle, always try to solve them by using trigonometric formulas and identities. Also, try to remember the values of basic trigonometric functions with simple angles.

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