BITSAT 2024 Paper 2
Exam BITSAT
Year 2024
Paper Paper 2
Subject Mathematics
Chapter Trigonometric Ratios & Identities BITSAT 2024 · Paper 2 · Mathematics · Trigonometric Ratios & Identities
The sum of all values of x in [0,2 π] , for which x+ 2 x+ 3 x+ 4...
The sum of all values of
x x x in
[ 0 , 2 π ] [0,2 \pi] [ 0 , 2 π ] , for which
sin x + sin 2 x + sin 3 x + sin 4 x = 0 \sin x+\sin 2 x+\sin 3 x+\sin 4 x=0 sin x + sin 2 x + sin 3 x + sin 4 x = 0 , is equal to :
Options A 8 π 8 \pi 8 π B 11 π 11 \pi 11 π C 12 π 12 \pi 12 π D 9 π 9 \pi 9 π Correct answer D. 9 π 9 \pi 9 π
Step-by-step solution
( sin x + sin 4 x ) + ( sin 2 x + sin 3 x ) = 0 ⇒ 2 sin 5 x 2 ⋅ cos 3 x 2 + 2 sin 5 x 2 ⋅ cos x 2 = 0 ⇒ 2 sin 5 x 2 { cos 3 x 2 + cos x 2 } = 0 ⇒ 2 sin 5 x 2 { 2 cos x cos x 2 } = 02 sin 5 x 2 = 0 ⇒ 5 x 2 = 0 , π , 2 π , 3 π , 4 π , 5 π , … ⇒ x = 0 , 2 π 5 , 4 π 5 , 6 π 5 , 8 π 5 , 2 π cos x 2 = 0 ⇒ x 2 = π 2 ⇒ x = π ; cos x = 0 ⇒ x = π 2 , 3 π 2 (\sin x+\sin 4 x)+(\sin 2 x+\sin 3 x)=0\Rightarrow 2 \sin \frac{5 x}{2} \cdot \cos \frac{3 x}{2}+2 \sin \frac{5 x}{2} \cdot \cos \frac{x}{2}=0\Rightarrow 2 \sin \frac{5 x}{2}\left\{\cos \frac{3 x}{2}+\cos \frac{x}{2}\right\}=0\Rightarrow 2 \sin \frac{5 x}{2}\left\{2 \cos x \cos \frac{x}{2}\right\}=02 \sin \frac{5 x}{2}=0 \Rightarrow \frac{5 x}{2}=0, \pi, 2 \pi, 3 \pi, 4 \pi, 5 \pi, \ldots\Rightarrow x=0, \frac{2 \pi}{5}, \frac{4 \pi}{5}, \frac{6 \pi}{5}, \frac{8 \pi}{5}, 2 \pi\cos \frac{x}{2}=0 \Rightarrow \frac{x}{2}=\frac{\pi}{2} \Rightarrow x=\pi ;\cos x=0 \Rightarrow x=\frac{\pi}{2}, \frac{3 \pi}{2} ( sin x + sin 4 x ) + ( sin 2 x + sin 3 x ) = 0 ⇒ 2 sin 2 5 x ⋅ cos 2 3 x + 2 sin 2 5 x ⋅ cos 2 x = 0 ⇒ 2 sin 2 5 x { cos 2 3 x + cos 2 x } = 0 ⇒ 2 sin 2 5 x { 2 cos x cos 2 x } = 02 sin 2 5 x = 0 ⇒ 2 5 x = 0 , π , 2 π , 3 π , 4 π , 5 π , … ⇒ x = 0 , 5 2 π , 5 4 π , 5 6 π , 5 8 π , 2 π cos 2 x = 0 ⇒ 2 x = 2 π ⇒ x = π ; cos x = 0 ⇒ x = 2 π , 2 3 π
So, sum
= 6 π + π + 2 π = 9 π =6 \pi+\pi+2 \pi=9 \pi = 6 π + π + 2 π = 9 π
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