Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

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

seo-qna
Last updated date: 29th Mar 2024
Total views: 420k
Views today: 13.20k
MVSAT 2024
Answer
VerifiedVerified
420k+ 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.