Sum

Solve the following equation:

\[\cos x + \cos 2x + \cos 3x = 0\]

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#### Solution

\[\cos x + \cos 2x + \cos 3x = 0\]

Now,

\[(\cos x + \cos3x) + \cos2x = 0\]

\[ \Rightarrow 2 \cos \left( \frac{4x}{2} \right) \cos \left( \frac{2x}{2} \right) + \cos2x = 0\]

\[ \Rightarrow 2 \cos2x \cos x + \cos2x = 0\]

\[ \Rightarrow \cos2x ( 2 \cos x + 1) = 0\]

\[\Rightarrow \cos 2x = 0\] or,

\[2 \cos x + 1 = 0\]

\[\Rightarrow \cos 2x = \cos \frac{\pi}{2}\] or

\[\cos x = - \frac{1}{2} = \cos \frac{2\pi}{3}\]

\[\Rightarrow 2x = (2n + 1) \frac{\pi}{2}\],

\[n \in Z\] or

\[x = 2m\pi \pm \frac{2\pi}{3}, m \in Z\]

\[\Rightarrow x = (2n + 1)\frac{\pi}{4}, n \in Z\]

\[x = 2m\pi \pm \frac{2\pi}{3}, m \in Z\]

Concept: Trigonometric Equations

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