The sum of all values of x in [0, 2$\pi$], for which sin x + sin 2x + sin 3x + sin 4x = 0, is equal to :
Solution
$(\sin x + \sin 4x) + (\sin 2x + \sin 3x) = 0$<br><br>$$ \Rightarrow 2\sin {{5x} \over 2}\left\{ {\cos {{3x} \over 2} + \cos {x \over 2}} \right\} = 0$$<br><br>$\Rightarrow 2\sin {{5x} \over 2}\left\{ {2\cos x\cos {x \over 2}} \right\} = 0$<br><br>$$2\sin {{5x} \over 2} = 0 \Rightarrow {{5x} \over 2} = 0,\pi ,2\pi ,3\pi ,4\pi ,5\pi $$<br><br>$$ \Rightarrow x = 0,{{2\pi } \over 5},{{4\pi } \over 5},{{6\pi } \over 5},{{8\pi } \over 5},2\pi $$<br><br>$$\cos {x \over 2} = 0 \Rightarrow {x \over 2} = {\pi \over 2} \Rightarrow x = \pi $$<br><br>$\cos x = 0 \Rightarrow x = {\pi \over 2},{{3\pi } \over 2}$<br><br>So, sum $= 6\pi + \pi + 2\pi = 9\pi$
About this question
Subject: Mathematics · Chapter: Trigonometric Functions · Topic: Trigonometric Ratios and Identities
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