If 1 6sin x-cos x-tan x are in G-P then

The correct option is B 2nπ±π3,n ∈ Z cos ^ 2 x = 1 6 sin x tan x 6cos ^ 2 x =sin ^ 2 x #160; cos x #160; 6cos ^ 3 x =1 cos ^ ...

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Byju's Answer

Standard XII

Mathematics

Variable Separable Method

If 1 6sin x...

Question

If $\frac{1}{6} sin x, cos x, tan x$ are in G.P then

n π \pm \frac{π}{3}, n \in Z

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2 n π \pm \frac{π}{3}, n \in Z

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n π \pm (- 1)^{n} \frac{π}{3}, n \in Z

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n π \pm \frac{π}{6}, n \in Z

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Solution

The correct option is B $2 n π \pm \frac{π}{3}, n \in Z$
${cos}^{2} x = \frac{1}{6} sin x tan x 6 {cos}^{2} x = \frac{{sin}^{2} x}{cos x} 6 {cos}^{3} x = 1 - {cos}^{2} x 6 {cos}^{3} x + {cos}^{2} x = 1 ∴ cos x = \frac{1}{2} ∴ cos x = cos \frac{π}{3} ∴ x = 2 n π \pm \frac{π}{3}, n \in Z$

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Similar questions

Q. If

4 \sin^{2} θ = 1

, then the values of θ are
(a)

2 n π \pm \frac{π}{3}, n \in Z

(b)

n π \pm \frac{π}{3}, n \in Z

(c)

n π \pm \frac{π}{6}, n \in Z

(d)

2 n π \pm \frac{π}{6}, n \in Z

Q. If

4 \sin^{2} x = 1

, then the values of x are
(a)

2 n π \pm \frac{π}{3}, n \in Z

(b)

n π \pm \frac{π}{3}, n \in Z

(c)

n π \pm \frac{π}{6}, n \in Z

(d)

2 n π \pm \frac{π}{6}, n \in Z

Q. The general value of x satisfying the equation

\sqrt{3} \sin x + \cos x = \sqrt{3}

is given by
(a)

x = n π + {(- 1)}^{n} \frac{π}{4} + \frac{π}{3}, n \in Z

(b)

x = n π + {(- 1)}^{n} \frac{π}{3} + \frac{π}{6}, n \in Z

(c)

x = n π \pm \frac{π}{6}, n \in Z

(d)

x = n π \pm \frac{π}{3}, n \in Z

Q. If

\sqrt{3} \cos x + \sin x = \sqrt{2}

, then general value of x is
(a)

n π + {(- 1)}^{n} \frac{π}{4}, n \in Z

(b)

{(- 1)}^{n} \frac{π}{4} - \frac{π}{3}, n \in Z

(c)

n π + \frac{π}{4} - \frac{π}{3}, n \in Z

(d)

n π + {(- 1)}^{n} \frac{π}{4} - \frac{π}{3}, n \in Z

Q. If

\sqrt{3} \cos θ + \sin θ = \sqrt{2}

, then general value of θ is
(a)

n π + {(- 1)}^{n} \frac{π}{4}, n \in Z

(b)

{(- 1)}^{n} \frac{π}{4} - \frac{π}{3}, n \in Z

(c)

n π + \frac{π}{4} - \frac{π}{3}, n \in Z

(d)

n π + {(- 1)}^{n} \frac{π}{4} - \frac{π}{3}, n \in Z

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If 16sinx,cosx,tanx are in G.P then

The correct option is B 2nπ±π3,n∈Zcos2x=16sinxtanx6cos2x=sin2xcosx6cos3x=1−cos2x6cos3x+cos2x=1∴cosx=12∴cosx=cosπ3∴x=2nπ±π3,n∈Z

If $\frac{1}{6} sin x, cos x, tan x$ are in G.P then

The correct option is B $2 n π \pm \frac{π}{3}, n \in Z$
${cos}^{2} x = \frac{1}{6} sin x tan x 6 {cos}^{2} x = \frac{{sin}^{2} x}{cos x} 6 {cos}^{3} x = 1 - {cos}^{2} x 6 {cos}^{3} x + {cos}^{2} x = 1 ∴ cos x = \frac{1}{2} ∴ cos x = cos \frac{π}{3} ∴ x = 2 n π \pm \frac{π}{3}, n \in Z$