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Chapter 19: Problem 1811

The number of solution of \(\cos x+\cos 2 x+\cos 3 x+\cos 4 x=0\) \(\mathrm{x} \in[0,2 \pi]\) is (a) 4 (b) 5 (c) 6 (d) 7

Short Answer

Expert verified
The number of solutions of the equation \(\cos x + \cos 2x + \cos 3x + \cos 4x = 0\) in the interval \(x \in [0, 2\pi]\) is (a) 4.

Step by step solution

01

Simplify the equation

We are given the equation: \(\cos x + \cos 2x + \cos 3x + \cos 4x = 0\) We will use the identities: \(\cos 2x = 2\cos^2 x - 1\) and \(\cos 3x = 4\cos^3 x - 3\cos x\) to rewrite the equation in terms of \(\cos x\). Then we will solve for \(\cos x\). The equation becomes: \(\cos x + (2\cos^2 x - 1) + (4\cos^3 x - 3\cos x) + \cos 4x = 0\)
02

Apply the double angle formula

Next, we will apply the double angle formula for \(\cos 4x\): \(\cos 4x = 2\cos^2 2x - 1\) Since we already have an expression for \(\cos 2x\), we can write \(\cos 4x\) in terms of \(\cos x\) as follows: \(\cos 4x = 2(2\cos^2 x -1)^2 - 1\) The equation now becomes: \(\cos x + (2\cos^2 x - 1) + (4\cos^3 x - 3\cos x) + (2(2\cos^2 x -1)^2 - 1) = 0\)
03

Simplify and solve for \(\cos x\)

Now, combining like terms in the equation, we get: \(8\cos^4 x - 8\cos^2 x + 1 = 0\) Let \(u = \cos^2 x\). Then the equation becomes: \(8u^2 - 8u + 1 = 0\) Solving the quadratic equation for \(u\), we get the solutions \(u_1 = \frac{1}{4}\) and \(u_2 = \frac{1}{2}\). Now, we will find the values of \(x\) corresponding to each solution: For \(u_1 = \frac{1}{4}\), we have \(\cos^2 x = \frac{1}{4}\), which gives us \(x = \pm \frac{\pi}{3}, \pm \frac{5\pi}{3}\). For \(u_2 = \frac{1}{2}\), we have \(\cos^2 x = \frac{1}{2}\), which gives us \(x = \pm \frac{\pi}{4}, \pm \frac{3\pi}{4}\).
04

Determine the number of solutions in the given interval

Now, let's count how many of these solutions lie in the given interval \(x \in [0, 2\pi]\): The solutions in the interval are \(x = \frac{\pi}{3}, \frac{5\pi}{3}, \frac{\pi}{4}, \frac{3\pi}{4}\). Since there are 4 solutions in the given interval, the answer is (a) 4.

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