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Question: The sum of all \(x \in \left[ {0,\pi } \right]\) which satisfy the equation \(\sin x + \dfrac{1}{2}\...

The sum of all x∈[0,π]x \in \left[ {0,\pi } \right] which satisfy the equation sin⁡x+12cos⁡x=sin⁡2(x+π4)\sin x + \dfrac{1}{2}\cos x = {\sin ^2}\left( {x + \dfrac{\pi }{4}} \right)is
A. π6\dfrac{\pi }{6}
B. 5π6\dfrac{{5\pi }}{6}
C. π\pi
D. 2π2\pi

Explanation

Solution

Hint: Use simple trigonometric formulas.

As we know that,
cos⁡2x=1−2sin⁡2x ∴sin⁡2x=1−cos⁡2x2  \cos 2x = 1 - 2{\sin ^2}x \\\ \therefore {\sin ^2}x = \dfrac{{1 - \cos 2x}}{2} \\\
Given: sin⁡x+12cos⁡x=sin⁡2(x+π4)\sin x + \dfrac{1}{2}\cos x = {\sin ^2}\left( {x + \dfrac{\pi }{4}} \right)
Substituting the value of sin⁡x\sin x, we get
⇒sin⁡x+12cos⁡x=12(1−cos⁡(2x+π2)) ⇒sin⁡x+12cos⁡x=12(1+sin⁡2x) (∵cos⁡(π2+θ)=−sin⁡θ) ⇒2sin⁡x+cos⁡x2=1+sin⁡2x2 ⇒2sin⁡x+cos⁡x=1+sin⁡2x  \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=2sin⁡xcos⁡x\sin 2x = 2\sin x\cos x in above equation, we get
⇒2sin⁡x+cos⁡x=1+2sin⁡xcos⁡x ⇒2sin⁡xcos⁡x−2sin⁡x−cos⁡x+1=0  \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 2sin⁡x2\sin x common, we get
⇒2sin⁡x(cos⁡x−1)−1(cos⁡x−1)=0 ⇒(2sin⁡x−1)(cos⁡x−1)=0  \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 (2sin⁡x−1)=0\left( {2\sin x - 1} \right) = 0 or (cos⁡x−1)=0\left( {\cos x - 1} \right) = 0.
If we take (2sin⁡x−1)=0\left( {2\sin x - 1} \right) = 0, then
⇒(2sin⁡x−1)=0 ⇒2sin⁡x=1 ⇒sin⁡x=12 ⇒x=π6,5π6  \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 (cos⁡x−1)=0\left( {\cos x - 1} \right) = 0, then
⇒(cos⁡x−1)=0 ⇒cos⁡x=1 ⇒x=0∘  \Rightarrow \left( {\cos x - 1} \right) = 0 \\\ \Rightarrow \cos x = 1 \\\ \Rightarrow x = {0^ \circ } \\\
Now, x∈[0,π]x \in \left[ {0,\pi } \right]is given.
So, the sum of all xx is.

x=π6+5π6+0 x=6π6 x=π  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.