The general solution of the equation sin 2 x -4 cos x isA- -2- 3 -2B- x- x-2 n -2- n - Z C- x - x - n -2-2- n - Z D- x - x - n - A -4- n - Z

The correct option is B {x:x= 2nπ±π2, n∈ Z} nbsp;sin 2x=4cos x ⇒ 2sin xcos x 4cos x=0 ⇒ 2cos x(sin x 2)=0 ⇒cos x=0 ⇒cos x= cosπ2 ⇒ x= 2nπ±π2, n∈ Z ...

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Standard XII

Mathematics

Properties Derived from Trigonometric Identities

The general s...

Question

The general solution of the equation $sin 2 x = 4 cos x$ is

{\frac{π}{2}, \frac{3 π}{2}}

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{x : x = 2 n π \pm \frac{π}{2}, n \in Z}

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{x : x = n π \pm \frac{π}{2}, n \in Z}

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{x : x = n π \pm \frac{π}{4}, n \in Z}

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Solution

The correct option is B ${x : x = 2 n π \pm \frac{π}{2}, n \in Z}$
$sin 2 x = 4 cos x$
$\Rightarrow 2 sin x cos x - 4 cos x = 0$
$\Rightarrow 2 cos x (sin x - 2) = 0$
$\Rightarrow cos x = 0$
$\Rightarrow cos x = cos \frac{π}{2}$
$\Rightarrow x = 2 n π \pm \frac{π}{2}, n \in Z$

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The general solution of the equation sin2x=4cosx is

The correct option is B {x:x=2nπ±π2,n∈Z} sin2x=4cosx ⇒2sinxcosx−4cosx=0 ⇒2cosx(sinx−2)=0 ⇒cosx=0 ⇒cosx=cosπ2 ⇒x=2nπ±π2,n∈Z

The general solution of the equation $sin 2 x = 4 cos x$ is

The correct option is B ${x : x = 2 n π \pm \frac{π}{2}, n \in Z}$
$sin 2 x = 4 cos x$
$\Rightarrow 2 sin x cos x - 4 cos x = 0$
$\Rightarrow 2 cos x (sin x - 2) = 0$
$\Rightarrow cos x = 0$
$\Rightarrow cos x = cos \frac{π}{2}$
$\Rightarrow x = 2 n π \pm \frac{π}{2}, n \in Z$