English

Find the General Solution of the Following Equation: Sin 2 X = √ 3 2

Advertisements
Advertisements

Question

Find the general solution of the following equation:

\[\sin 2x = \frac{\sqrt{3}}{2}\]
Sum
Advertisements

Solution

We have:

\[\sin2x = \frac{\sqrt{3}}{2}\]

⇒ \[\sin2x = \sin \frac{\pi}{3}\]

⇒ \[2x = n\pi + ( - 1 )^n \frac{\pi}{3}\]

\[n \in Z\]

⇒ \[x = \frac{n\pi}{2} + ( - 1 )^n \frac{\pi}{6}\],

\[n \in Z\]
shaalaa.com
  Is there an error in this question or solution?
Chapter 11: Trigonometric equations - Exercise 11.1 [Page 21]

APPEARS IN

R.D. Sharma Mathematics [English] Class 11
Chapter 11 Trigonometric equations
Exercise 11.1 | Q 2.01 | Page 21

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

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


Find the principal and general solutions of the equation sec x = 2


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


Find the general solution of the equation cos 4 x = cos 2 x


Find the general solution for each of the following equations sec2 2x = 1– tan 2x


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  \[2 T_6 - 3 T_4 + 1 = 0\]


Prove that:

\[3\sin\frac{\pi}{6}\sec\frac{\pi}{3} - 4\sin\frac{5\pi}{6}\cot\frac{\pi}{4} = 1\]

 


Prove that:
\[\frac{\cos (2\pi + x) cosec (2\pi + x) \tan (\pi/2 + x)}{\sec(\pi/2 + x)\cos x \cot(\pi + x)} = 1\]

 


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

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

 


In a ∆ABC, prove that:

\[\tan\frac{A + B}{2} = \cot\frac{C}{2}\]

Prove that:
\[\sin \frac{13\pi}{3}\sin\frac{2\pi}{3} + \cos\frac{4\pi}{3}\sin\frac{13\pi}{6} = \frac{1}{2}\]


Prove that:

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

 


If tan x = \[x - \frac{1}{4x}\], then sec x − tan x is equal to


If x = r sin θ cos ϕ, y = r sin θ sin ϕ and r cos θ, then x2 + y2 + z2 is independent of


If \[cosec x - \cot x = \frac{1}{2}, 0 < x < \frac{\pi}{2},\]

 

The value of sin25° + sin210° + sin215° + ... + sin285° + sin290° is


sin2 π/18 + sin2 π/9 + sin2 7π/18 + sin2 4π/9 =


If tan A + cot A = 4, then tan4 A + cot4 A is equal to


If tan θ + sec θ =ex, then cos θ equals


If sec x + tan x = k, cos x =


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

 

Find the general solution of the following equation:

\[\tan x = - \frac{1}{\sqrt{3}}\]

Find the general solution of the following equation:

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

Find the general solution of the following equation:

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

Find the general solution of the following equation:

\[\tan 3x = \cot x\]

Find the general solution of the following equation:

\[\tan mx + \cot nx = 0\]

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


Solve the following equation:

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

Solve the following equation:

\[\sin 3x - \sin x = 4 \cos^2 x - 2\]

Solve the following equation:

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

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


Write the number of points of intersection of the curves

\[2y = - 1 \text{ and }y = cosec x\]

If \[3\tan\left( x - 15^\circ \right) = \tan\left( x + 15^\circ \right)\] \[0 < x < 90^\circ\], find θ.


Solve the following equations for which solution lies in the interval 0° ≤ θ < 360°

2 sin2x + 1 = 3 sin x


Solve the following equations:
2cos 2x – 7 cos x + 3 = 0


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×