हिंदी

Find the General Solution of the Following Equation: Sin 9 X = Sin X - Mathematics

Advertisements
Advertisements

प्रश्न

Find the general solution of the following equation:

\[\sin 9x = \sin x\]
योग
Advertisements

उत्तर

We have:

\[\sin9x = \sin x\]

⇒ \[\sin9x - \sin x = 0\]

⇒ \[2 \sin \left( \frac{9x - x}{2} \right) \cos \left( \frac{9x + x}{2} \right) = 0\]

⇒ \[\sin \frac{8x}{2} = 0\] or \[\cos \frac{10x}{2} = 0\]

⇒ \[\sin 4x = 0\] or \[\cos 5x = 0\]

⇒ \[4x = n\pi\]

\[n \in Z\] or \[5x = (2n + 1)\frac{\pi}{2}\],

\[n \in Z\]

⇒ \[x = \frac{n\pi}{4}\],

\[n \in Z\] or \[x = (2n + 1)\frac{\pi}{10}\],

\[n \in Z\]
shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 11: Trigonometric equations - Exercise 11.1 [पृष्ठ २१]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
अध्याय 11 Trigonometric equations
Exercise 11.1 | Q 2.03 | पृष्ठ २१

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

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 general solution of the equation cos 3x + cos x – cos 2x = 0


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


If \[\tan x = \frac{b}{a}\] , then find the values of \[\sqrt{\frac{a + b}{a - b}} + \sqrt{\frac{a - b}{a + b}}\].


Prove the:
\[ \sqrt{\frac{1 - \sin x}{1 + \sin x}} + \sqrt{\frac{1 + \sin x}{1 - \sin x}} = - \frac{2}{\cos x},\text{ where }\frac{\pi}{2} < x < \pi\]


Prove that: cos 24° + cos 55° + cos 125° + cos 204° + cos 300° = \[\frac{1}{2}\]


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


In a ∆ABC, prove that:

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

Prove that:
\[\tan 4\pi - \cos\frac{3\pi}{2} - \sin\frac{5\pi}{6}\cos\frac{2\pi}{3} = \frac{1}{4}\]


Prove that:

\[\sin\frac{10\pi}{3}\cos\frac{13\pi}{6} + \cos\frac{8\pi}{3}\sin\frac{5\pi}{6} = - 1\]

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


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

 

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

 


\[\sec^2 x = \frac{4xy}{(x + y )^2}\] is true if and only if

 


If x sin 45° cos2 60° = \[\frac{\tan^2 60^\circ cosec30^\circ}{\sec45^\circ \cot^{2^\circ} 30^\circ}\], then x =

 

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

 

Find the general solution of the following equation:

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

Solve the following equation:

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

Solve the following equation:

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

Solve the following equation:
\[\sin x - 3\sin2x + \sin3x = \cos x - 3\cos2x + \cos3x\]


Solve the following equation:
 sin x tan x – 1 = tan x – sin x

 


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


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


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

 


Find the principal solution and general solution of the following:
cot θ = `sqrt(3)`


Solve the following equations:
sin 5x − sin x = cos 3


Solve the following equations:
2 cos2θ + 3 sin θ – 3 = θ


Solve the following equations:
sin 2θ – cos 2θ – sin θ + cos θ = θ


Choose the correct alternative:
If tan 40° = λ, then `(tan 140^circ - tan 130^circ)/(1 + tan 140^circ *  tan 130^circ)` =


Solve 2 tan2x + sec2x = 2 for 0 ≤ x ≤ 2π.


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×