The solution set of cos 2 x-2 -3sin x-4 cos x-5-0 is

The correct option is D ϕ nbsp; nbsp;cos^2x 2 √(3)sin x 4 cos x+5=0 ⇒cos^2x 2 √(3)sin x 4 cos x+3(cos^2x + sin^2 x)+2=0 ⇒ (4 cos^2 x 4 cos x+1)+(3sin^2 ...

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Standard XII

Mathematics

General Solution of Trigonometric Equation

The solution ...

Question

The solution set of ${cos}^{2} x - 2 \sqrt{3} sin x - 4 cos x + 5 = 0$ is

{2 n π \pm \frac{π}{6}}

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{2 n π \pm \frac{π}{6}}

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{2 n π \pm \frac{π}{6}}

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ϕ

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Solution

The correct option is D $ϕ$
${cos}^{2} x - 2 \sqrt{3} sin x - 4 cos x + 5 = 0 \Rightarrow {cos}^{2} x - 2 \sqrt{3} sin x - 4 cos x + 3 ({cos}^{2} x + {sin}^{2} x) + 2 = 0 \Rightarrow (4 {cos}^{2} x - 4 cos x + 1) + (3 {sin}^{2} x - 2 \sqrt{3} sin x + 1) = 0 \Rightarrow (2 cos x - 1)^{2} + (\sqrt{3} sin x - 1)^{2} = 0 \Rightarrow cos x = \frac{1}{2} and sin x = \frac{1}{\sqrt{3}}$
$\Rightarrow$ Not possible

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Similar questions

3 sin x + 4 cos x = 5 \Rightarrow 6 tan \frac{x}{2} - 9 {tan}^{2} \frac{x}{2} =

Q. Solve

3 sin x + 4 cos x = 5 \Rightarrow 6 tan \frac{x}{2} - 9 {tan}^{2} \frac{x}{2} =

Q. Let

f

be an odd function defined on the real numbers such that

f (x) = 3 sin x + 4 cos x

, for

x \geq 0

, then

f (x)

for

x < 0

3 sin x + 4 cos x = 5

, then the value of

3 tan \frac{x}{2}

Q. The Range of

f (x) = 3 sin x + 4 cos x + 5

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The solution set of cos2x−2√3sinx−4cosx+5=0 is

The correct option is D ϕ cos2x−2√3sinx−4cosx+5=0⇒cos2x−2√3sinx−4cosx+3(cos2x+sin2x)+2=0⇒(4cos2x−4cosx+1)+(3sin2x−2√3sinx+1)=0⇒(2cosx−1)2+(√3sinx−1)2=0⇒cosx=12 and sinx=1√3 ⇒Not possible

The solution set of ${cos}^{2} x - 2 \sqrt{3} sin x - 4 cos x + 5 = 0$ is