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Question

Write the solution set of the equation $\left(2\mathrm{cos}x+1\right)\left(4\mathrm{cos}x+5\right)=0$ in the interval [0, 2π].

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Solution

Given: $\left(2\mathrm{cos}x+1\right)\left(4\mathrm{cos}x+5\right)=0$ Now, $2\mathrm{cos}x+1=0$ or $4\mathrm{cos}x+5=0$ $⇒\mathrm{cos}x=-\frac{1}{2}$ or $\mathrm{cos}x=-\frac{5}{4}$ $\mathrm{cos}x=-\frac{5}{4}$ is not possible. Thus, we have: $\mathrm{cos}x=-\frac{1}{2}\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}x=\mathrm{cos}\frac{2\mathrm{\pi }}{3}\phantom{\rule{0ex}{0ex}}⇒x=2n\mathrm{\pi }±\frac{2\mathrm{\pi }}{3}$ By putting n = 0 and n = 1 in the above equation, we get: $x=\frac{2\mathrm{\pi }}{3}$ or $x=\frac{4\mathrm{\pi }}{3}$ in the interval $\left[0,2\mathrm{\pi }\right]$ For the other value of n, x will not satisfy the given condition. ∴ $x=\frac{2\mathrm{\pi }}{3}$ and $\frac{4\mathrm{\pi }}{3}$

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