Question 32

The sum of all solutions of the equation $$4 \sin^2 x - 4 \cos x = 1$$ in the interval $$[0, 2\pi]$$ is

Solution

$$4 \sin^2 x - 4 \cos x = 1$$

=> $$4\left(1-\cos\ ^2x\right)-4\cos x-1=0$$

=> $$4\cos\ ^2x+4\cos x+1=4$$

$$\left(2\cos x+1\right)^2=2^2$$

$$2\cos x+1=\pm\ 2$$

$$2\cos x=1\ or\ 2\cos x=-3$$

$$\cos x=\frac{1}{2}\ or\ \cos x=-\frac{3}{2}$$

-3/2 is not possible

Hence x can be $$\frac{\pi}{3}\ or\ 2\pi\ -\frac{\pi}{3}$$

Sum = $$2\pi$$

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