X - R- cos 2 x-2 cos 2 x-2 is equal to

The correct option is D nπ±π/6: n∈ Ζ Given equation is cos2x+2cos^2x=2 ⇒ 2cos^2x 1+2cos^2x=2 ⇒ 4cos^2x=3 ⇒ cos^2x=3/4 ⇒ cos^2x=cos^2π/6 ⇒ x=nπ±π/6:n∈ Z ...

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Byju's Answer

Standard XI

Mathematics

Solving Simultaneous Trigonometric Equations

x ∈ R: cos 2 ...

Question

${x \in R : c o s 2 x + 2 c o s^{2} x = 2}$ is equal to

2nπ + π/3: n∈ Ζ

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nπ±π/6: n∈ Ζ

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nπ + π/3: n∈ Ζ

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2nπ-π/3: n∈ Ζ

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Solution

The correct option is B
nπ±π/6: n∈ Ζ

Given equation is

$c o s 2 x + 2 c o s^{2} x = 2 \Rightarrow 2 c o s^{2} x - 1 + 2 c o s^{2} x = 2 \Rightarrow 4 c o s^{2} x = 3 \Rightarrow c o s^{2} x = \frac{3}{4} \Rightarrow c o s^{2} x = c o s^{2} \frac{π}{6} \Rightarrow x = n π \pm \frac{π}{6} : n \in Z$

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Similar questions

{x \in R : cos 2 x + 2 {cos}^{2} x = 2}

is equal to

2 s i n^{2} ((\frac{π}{2}) c o s^{2} x) = 1 - c o s (π s i n 2 x), x \neq (2 n + 1) \frac{π}{2}, n ϵ I

, then

c o s 2 x

is equal to

Q. if

2 {sin}^{2} ((\frac{π}{2}) {cos}^{2} x) = 1 - cos (π {sin}^{2} 2 x), x \neq (2 n + 1) \frac{π}{2}, n ϵ I

,then

cos 2 x

is equal to

Q. If

2 sin^{2} ((\frac{π}{2}) cos^{2} x) = 1 - cos (π sin 2 x); x \neq \frac{(2 n + 1) π}{2}, n \in I

, then

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Q. If

g (x) = {(4 {cos}^{4} x - 2 cos 2 x - \frac{cos 4 x}{2} - x^{7})}^{1 / 7}

for

x \in R,

then the value of

g (g (100))

is equal to

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{x∈R:cos2x+2cos2x=2} is equal to

The correct option is B nπ±π/6: n∈ Ζ Given equation is cos2x+2cos2x=2⇒2cos2x−1+2cos2x=2⇒4cos2x=3⇒cos2x=34⇒cos2x=cos2π6⇒x=nπ±π6:n∈Z

${x \in R : c o s 2 x + 2 c o s^{2} x = 2}$ is equal to