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Find the General Solution of the Equation Sin 2x + Cos X = 0 - Mathematics

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प्रश्न

Find the general solution of the equation sin 2x + cos x = 0

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उत्तर

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पाठ 3: Trigonometric Functions - Exercise 3.4 [पृष्ठ ७८]

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एनसीईआरटी Mathematics [English] Class 11
पाठ 3 Trigonometric Functions
Exercise 3.4 | Q 7 | पृष्ठ ७८

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

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Prove that

\[\frac{cosec(90^\circ + x) + \cot(450^\circ + x)}{cosec(90^\circ - x) + \tan(180^\circ - x)} + \frac{\tan(180^\circ + x) + \sec(180^\circ - x)}{\tan(360^\circ + x) - \sec( - x)} = 2\]

 


Prove that

\[\left\{ 1 + \cot x - \sec\left( \frac{\pi}{2} + x \right) \right\}\left\{ 1 + \cot x + \sec\left( \frac{\pi}{2} + x \right) \right\} = 2\cot x\]

 


In a ∆ABC, prove that:
cos (A + B) + cos C = 0


Find x from the following equations:
\[cosec\left( \frac{\pi}{2} + \theta \right) + x \cos \theta \cot\left( \frac{\pi}{2} + \theta \right) = \sin\left( \frac{\pi}{2} + \theta \right)\]


Prove that:

\[\tan\frac{5\pi}{4}\cot\frac{9\pi}{4} + \tan\frac{17\pi}{4}\cot\frac{15\pi}{4} = 0\]

 


If \[\frac{\pi}{2} < x < \frac{3\pi}{2},\text{ then }\sqrt{\frac{1 - \sin x}{1 + \sin x}}\] is equal to

 


\[\sqrt{\frac{1 + \cos x}{1 - \cos x}}\] is equal to

 


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The value of \[\tan1^\circ \tan2^\circ \tan3^\circ . . . \tan89^\circ\] is

 

Find the general solution of the following equation:

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Find the general solution of the following equation:

\[\cos x = - \frac{\sqrt{3}}{2}\]

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\[\cos 3x = \frac{1}{2}\]

Solve the following equation:

\[2 \sin^2 x + \sqrt{3} \cos x + 1 = 0\]

Solve the following equation:

\[4 \sin^2 x - 8 \cos x + 1 = 0\]

Solve the following equation:

\[3 \cos^2 x - 2\sqrt{3} \sin x \cos x - 3 \sin^2 x = 0\]

Solve the following equation:

\[\tan 3x + \tan x = 2\tan 2x\]

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Solve the following equation:
\[2^{\sin^2 x} + 2^{\cos^2 x} = 2\sqrt{2}\]


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Choose the correct alternative:
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