हिंदी

Write the General Solutions of Tan2 2x = 1.

Advertisements
Advertisements

प्रश्न

Write the general solutions of tan2 2x = 1.

 
योग
Advertisements

उत्तर

Given: 
\[\tan^2 2x = 1\]
\[ \Rightarrow \tan 2x = \tan \frac{\pi}{4}\]
\[ \Rightarrow 2x = n\pi + \frac{\pi}{4}\]
\[ \Rightarrow x = \frac{n\pi}{2} + \frac{\pi}{8}, n \in Z\]
Hence, the general solution of the equation is 
\[\frac{n\pi}{2} + \frac{\pi}{8}, n \in Z .\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 11: Trigonometric equations - Exercise 11.2 [पृष्ठ २६]

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 11 Trigonometric equations
Exercise 11.2 | Q 3 | पृष्ठ २६

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

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


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


Find the general solution of the equation  sin x + sin 3x + sin 5x = 0


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


Prove that

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

 


Prove that:

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

 


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

 


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


If tan \[x = - \frac{1}{\sqrt{5}}\] and θ lies in the IV quadrant, then the value of cos x is

 

If \[\frac{3\pi}{4} < \alpha < \pi, \text{ then }\sqrt{2\cot \alpha + \frac{1}{\sin^2 \alpha}}\] is equal to


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

 


If x is an acute angle and \[\tan x = \frac{1}{\sqrt{7}}\], then the value of \[\frac{{cosec}^2 x - \sec^2 x}{{cosec}^2 x + \sec^2 x}\] is


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


If A lies in second quadrant 3tan A + 4 = 0, then the value of 2cot A − 5cosA + sin A is equal to


The value of \[\cos1^\circ \cos2^\circ \cos3^\circ . . . \cos179^\circ\] is

 

Find the general solution of the following equation:

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

Find the general solution of the following equation:

\[cosec x = - \sqrt{2}\]

Find the general solution of the following equation:

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

Find the general solution of the following equation:

\[\sin 2x = \cos 3x\]

Find the general solution of the following equation:

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

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


Solve the following equation:

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

Solve the following equation:

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

Solve the following equation:

\[\sin x + \cos x = 1\]

Solve the following equation:
\[\sec x\cos5x + 1 = 0, 0 < x < \frac{\pi}{2}\]


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


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


Write the number of solutions of the equation tan x + sec x = 2 cos x in the interval [0, 2π].


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


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


Find the principal solution and general solution of the following:
tan θ = `- 1/sqrt(3)`


Solve the following equations:
`sin theta + sqrt(3) cos theta` = 1


Choose the correct alternative:
If tan α and tan β are the roots of x2 + ax + b = 0 then `(sin(alpha + beta))/(sin alpha sin beta)` is equal to


Choose the correct alternative:
`(cos 6x + 6 cos 4x + 15cos x + 10)/(cos 5x + 5cs 3x + 10 cos x)` is equal to


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


If a cosθ + b sinθ = m and a sinθ - b cosθ = n, then show that a2 + b2 = m2 + n2 


Find the general solution of the equation sinx – 3sin2x + sin3x = cosx – 3cos2x + cos3x


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×