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Question Number 127814 by dactor last updated on 02/Jan/21

	Answered by mnjuly1970 last updated on 02/Jan/21
	$solution 2cos^2 (x)−1−cos(x)+1=0 ⇒cos(x)(2cos(x)−1)=0 cos(x)=0⇒ x=2kπ ; k∈Z cos(x)=(1/2) ⇒{_(x=2kπ−(π/3)) ^(x=2kπ+(π/3)) k∈Z$
	$s o l u t i o n$ $2 {c o s}^{2} (x) - 1 - c o s (x) + 1 = 0$ $\Rightarrow c o s (x) (2 c o s (x) - 1) = 0$ $c o s (x) = 0 \Rightarrow x = 2 k π; k \in Z$ $c o s (x) = \frac{1}{2} \Rightarrow {_{x = 2 k π - \frac{π}{3}}^{x = 2 k π + \frac{π}{3}} k \in Z$
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