मराठी

Find the General Solution of the Following Equation: √ 3 Sec X = 2 - Mathematics

Advertisements
Advertisements

प्रश्न

Find the general solution of the following equation:

\[\sqrt{3} \sec x = 2\]
बेरीज
Advertisements

उत्तर

We have:

\[\sqrt{3} \sec x = 2\]
⇒ \[\sec x = \frac{2}{\sqrt{3}}\] (or) 
\[\cos x = \frac{\sqrt{3}}{2}\]
The value of x satisfying \[\cos x = \frac{\sqrt{3}}{2}\] is \[\frac{\pi}{6}\].
∴ \[\cos x = \frac{\sqrt{3}}{2}\]
⇒ \[\cos x = \cos\frac{\pi}{6}\]
⇒ \[x = 2n\pi \pm \frac{\pi}{6}\],
\[n \in Z\]
shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 11: Trigonometric equations - Exercise 11.1 [पृष्ठ २१]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
पाठ 11 Trigonometric equations
Exercise 11.1 | Q 1.6 | पृष्ठ २१

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

Find the principal and general solutions of the equation  `cot x = -sqrt3`


Find the general solution of cosec x = –2


If \[\cot x \left( 1 + \sin x \right) = 4 m \text{ and }\cot x \left( 1 - \sin x \right) = 4 n,\] \[\left( m^2 + n^2 \right)^2 = mn\]


If \[a = \sec x - \tan x \text{ and }b = cosec x + \cot x\], then shown that  \[ab + a - b + 1 = 0\]


If \[T_n = \sin^n x + \cos^n x\], prove that \[6 T_{10} - 15 T_8 + 10 T_6 - 1 = 0\]


Prove that:  tan 225° cot 405° + tan 765° cot 675° = 0


Prove that cos 570° sin 510° + sin (−330°) cos (−390°) = 0

Prove that: \[\tan\frac{11\pi}{3} - 2\sin\frac{4\pi}{6} - \frac{3}{4} {cosec}^2 \frac{\pi}{4} + 4 \cos^2 \frac{17\pi}{6} = \frac{3 - 4\sqrt{3}}{2}\]

 


Prove that

\[\frac{\sin(180^\circ + x) \cos(90^\circ + x) \tan(270^\circ - x) \cot(360^\circ - x)}{\sin(360^\circ - x) \cos(360^\circ + x) cosec( - x) \sin(270^\circ + x)} = 1\]

 


Prove that:
\[\sin^2 \frac{\pi}{18} + \sin^2 \frac{\pi}{9} + \sin^2 \frac{7\pi}{18} + \sin^2 \frac{4\pi}{9} = 2\]

 

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


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

 


If tan x + sec x = \[\sqrt{3}\], 0 < x < π, then x is equal to


If \[cosec x + \cot x = \frac{11}{2}\], then tan x =

 


If \[f\left( x \right) = \cos^2 x + \sec^2 x\], then


Which of the following is incorrect?


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

 

Find the general solution of the following equation:

\[\sin x = \frac{1}{2}\]

Find the general solution of the following equation:

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

Find the general solution of the following equation:

\[\sin 9x = \sin x\]

Solve the following equation:
\[\sin^2 x - \cos x = \frac{1}{4}\]


Solve the following equation:

\[\cos x + \cos 3x - \cos 2x = 0\]

Solve the following equation:

\[\cos x \cos 2x \cos 3x = \frac{1}{4}\]

Solve the following equation:

\[\sin x + \sin 2x + \sin 3 = 0\]

If secx cos5x + 1 = 0, where \[0 < x \leq \frac{\pi}{2}\], find the value of x.


If \[2 \sin^2 x = 3\cos x\]. where \[0 \leq x \leq 2\pi\], then find the value of x.


If \[\cos x + \sqrt{3} \sin x = 2,\text{ then }x =\]

 


A solution of the equation \[\cos^2 x + \sin x + 1 = 0\], lies in the interval


If \[4 \sin^2 x = 1\], then the values of x are

 


If \[\cot x - \tan x = \sec x\], then, x is equal to

 


A value of x satisfying \[\cos x + \sqrt{3} \sin x = 2\] is

 

The number of values of x in the interval [0, 5 π] satisfying the equation \[3 \sin^2 x - 7 \sin x + 2 = 0\] is


Find the principal solution and general solution of the following:
sin θ = `-1/sqrt(2)`


Solve the following equations:
cos θ + cos 3θ = 2 cos 2θ


Number of solutions of the equation tan x + sec x = 2 cosx lying in the interval [0, 2π] is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×