Courses
Courses for Kids
Free study material
Free LIVE classes
More
Questions & Answers

The sum of all $x \in \left[ {0,\pi } \right]$ which satisfy the equation $\sin x + \dfrac{1}{2}\cos x = {\sin ^2}\left( {x + \dfrac{\pi }{4}} \right)$isA. $\dfrac{\pi }{6}$B. $\dfrac{{5\pi }}{6}$C. $\pi$D. $2\pi$

Last updated date: 28th Mar 2023
Total views: 311.1k
Views today: 4.87k
Answer
Verified
311.1k+ views
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.