English

Solve the Following Equation: Tan X + Tan 2 X = Tan 3 X

Advertisements
Advertisements

Question

Solve the following equation:

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

Solution

Given:

\[\tan x + \tan2x = \tan3x\]
Now, 
\[\tan x + \tan2x = \tan (x + 2x)\]
\[ \Rightarrow \tan x + \tan 2x = \frac{\tan x + \tan2x}{1 - \tan x \tan2x}\]
\[ \Rightarrow \tan x + \tan2x - \frac{\tan x + \tan2x}{1 - \tan x \tan2x} = 0\]
\[ \Rightarrow (\tan x + \tan2x) (1 - \tan x \tan2x) - (\tan x + \tan2x) = 0\]
\[ \Rightarrow (\tan x + \tan 2x) (1 - \tan x \tan2x - 1) = 0\]
\[ \Rightarrow (\tan x + \tan2x) ( - \tan x \tan2x) = 0\]
\[\Rightarrow \tan x + \tan 2x = 0\] or
\[\tan x \tan2x = 0\]

Now,

\[\tan x + \tan 2x = 0 \]
\[ \Rightarrow \tan x = - \tan 2x\]
\[ \Rightarrow \tan x = \tan - 2x\]
\[ \Rightarrow x = n\pi - 2x, n \in Z\]
\[ \Rightarrow 3x = n\pi \]
\[ \Rightarrow x = \frac{n\pi}{3}, n \in Z\]

And,

\[\tan x + \tan 2x = 0 \]
\[ \Rightarrow \tan x = - \tan 2x\]
\[ \Rightarrow \tan x = \tan - 2x\]
\[ \Rightarrow x = n\pi - 2x, n \in Z\]
\[ \Rightarrow 3x = n\pi \]
\[ \Rightarrow x = \frac{n\pi}{3}, n \in Z\]

∴ \[x = \frac{n\pi}{3}, n \in Z\] or

\[x = \frac{n\pi}{3}, n \in Z\]
shaalaa.com
  Is there an error in this question or solution?
Chapter 11: Trigonometric equations - Exercise 11.1 [Page 22]

APPEARS IN

R.D. Sharma Mathematics [English] Class 11
Chapter 11 Trigonometric equations
Exercise 11.1 | Q 5.2 | Page 22

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

Find the general solution of cosec x = –2


If \[\tan x = \frac{a}{b},\] show that

\[\frac{a \sin x - b \cos x}{a \sin x + b \cos x} = \frac{a^2 - b^2}{a^2 + b^2}\]

If \[cosec x - \sin x = a^3 , \sec x - \cos x = b^3\], then prove that \[a^2 b^2 \left( a^2 + b^2 \right) = 1\]


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


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

 

In a ∆ABC, prove that:

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

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


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


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


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

 

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

 


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

 

Find the general solution of the following equation:

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

Find the general solution of the following equation:

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

Find the general solution of the following equation:

\[\tan px = \cot qx\]

 


Find the general solution of the following equation:

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

Solve the following equation:

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

Solve the following equation:

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

Solve the following equation:

\[\sin x + \sin 2x + \sin 3x + \sin 4x = 0\]

Solve the following equation:

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

Solve the following equation:

`cosec  x = 1 + cot x`


Solve the following equation:
\[2 \sin^2 x = 3\cos x, 0 \leq x \leq 2\pi\]


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


If cos x = k has exactly one solution in [0, 2π], then write the values(s) of k.

 

Write the values of x in [0, π] for which \[\sin 2x, \frac{1}{2}\]

 and cos 2x are in A.P.


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


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

2 sin2x + 1 = 3 sin x


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


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


Solve `sqrt(3)` cos θ + sin θ = `sqrt(2)`


If 2sin2θ = 3cosθ, where 0 ≤ θ ≤ 2π, then find the value of θ.


Find the general solution of the equation 5cos2θ + 7sin2θ – 6 = 0


The minimum value of 3cosx + 4sinx + 8 is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×