If $0 \leq x < 2 π$ , then the number of real values of x, which satisfy the equation $cos x + cos 2 x + cos 3 x + cos 4 x = 0$ , is

3

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5

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7

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9

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Solution

The correct option is C $7$
$Given equation is cos x + cos 2 x + cos 3 x + cos 4 x = 0 \Rightarrow (cos x + cos 3 x) + (cos 2 x + cos 4 x) = 0 \Rightarrow 2 cos 2 x cos x + 2 cos 3 x cos x = 0 \Rightarrow 2 cos x (cos 2 x + cos 3 x) = 0 \Rightarrow 2 cos x (2 cos \frac{5 x}{2} cos \frac{x}{2}) = 0 \Rightarrow cos x \cdot cos \frac{5 x}{2} \cdot cos \frac{x}{2} = 0 \Rightarrow cos x = 0 or cos \frac{5 x}{2} = 0 or cos \frac{x}{2} = 0 Now, cos x = 0 \Rightarrow x = \frac{π}{2}, \frac{3 π}{2} [∵ 0 \leq x < 2 π] cos \frac{5 x}{2} = 0 \Rightarrow \frac{5 x}{2} = \frac{π}{2}, \frac{3 π}{2}, \frac{5 π}{2}, \frac{7 π}{2}, \frac{9 π}{2}, \frac{11 π}{2}, . . . . \Rightarrow x = \frac{π}{5}, \frac{3 π}{5}, π, \frac{7 π}{5}, \frac{9 π}{5} [∵ 0 \leq x < 2 π] and cos \frac{x}{2} = 0 \Rightarrow \frac{x}{2} = \frac{π}{2}, \frac{3 π}{2}, \frac{5 π}{2}, . . . . . . \Rightarrow x = π [∵ 0 \leq x < 2 π] Hence, x = \frac{π}{2}, \frac{3 π}{2}, π, \frac{π}{5}, \frac{3 π}{5}, \frac{7 π}{5}, \frac{9 π}{5}$

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Similar questions

If $0 \leq x < 2 π$ , then the number of real values of x, which satisfy the equation cosx + cos2x + cos3x + cos4x = 0, is:

0 \leq x < 2 π

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2^{(1 + | cos x | + ∣ ∣ c o s^{2} x ∣ ∣ + ∣ ∣ c o s^{3} x ∣ ∣) + ∣ ∣ c o s^{4} x ∣ ∣ . . . . . . . . . . . . . . . \infty} = 4

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