Advertisements
Advertisements
प्रश्न
Solve the following equation:
Advertisements
उत्तर
Given:
\[ \Rightarrow \tan x + \tan 2x = \frac{\tan x + \tan2x}{1 - \tan x \tan2x}\]
\[ \Rightarrow \tan x + \tan2x - \frac{\tan x + \tan2x}{1 - \tan x \tan2x} = 0\]
\[ \Rightarrow (\tan x + \tan2x) (1 - \tan x \tan2x) - (\tan x + \tan2x) = 0\]
\[ \Rightarrow (\tan x + \tan 2x) (1 - \tan x \tan2x - 1) = 0\]
\[ \Rightarrow (\tan x + \tan2x) ( - \tan x \tan2x) = 0\]
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
APPEARS IN
संबंधित प्रश्न
Find the principal and general solutions of the equation `tan x = sqrt3`
If \[\sin x + \cos x = m\], then prove that \[\sin^6 x + \cos^6 x = \frac{4 - 3 \left( m^2 - 1 \right)^2}{4}\], where \[m^2 \leq 2\]
If \[T_n = \sin^n x + \cos^n x\], prove that \[2 T_6 - 3 T_4 + 1 = 0\]
Prove that:
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
In a ∆ABC, prove that:
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:
If \[0 < x < \frac{\pi}{2}\], and if \[\frac{y + 1}{1 - y} = \sqrt{\frac{1 + \sin x}{1 - \sin x}}\], then y 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 \[\frac{3\pi}{4} < \alpha < \pi, \text{ then }\sqrt{2\cot \alpha + \frac{1}{\sin^2 \alpha}}\] is equal to
If \[cosec x - \cot x = \frac{1}{2}, 0 < x < \frac{\pi}{2},\]
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 x sin 45° cos2 60° = \[\frac{\tan^2 60^\circ cosec30^\circ}{\sec45^\circ \cot^{2^\circ} 30^\circ}\], then x =
Which of the following is incorrect?
Which of the following is correct?
Find the general solution of the following equation:
Solve the following equation:
Solve the following equation:
Solve the following equation:
`cosec x = 1 + cot x`
Solve the following equation:
\[\sin x - 3\sin2x + \sin3x = \cos x - 3\cos2x + \cos3x\]
Write the general solutions of tan2 2x = 1.
Write the solution set of the equation
The smallest value of x satisfying the equation
The general value of x satisfying the equation
\[\sqrt{3} \sin x + \cos x = \sqrt{3}\]
The smallest positive angle which satisfies the equation
The equation \[3 \cos x + 4 \sin x = 6\] has .... solution.
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
Solve the following equations for which solution lies in the interval 0° ≤ θ < 360°
2 cos2x + 1 = – 3 cos x
Solve the following equations:
sin 2θ – cos 2θ – sin θ + cos θ = θ
Choose the correct alternative:
If cos pθ + cos qθ = 0 and if p ≠ q, then θ is equal to (n is any integer)
Choose the correct alternative:
If f(θ) = |sin θ| + |cos θ| , θ ∈ R, then f(θ) is in the interval
Solve `sqrt(3)` cos θ + sin θ = `sqrt(2)`
Number of solutions of the equation tan x + sec x = 2 cosx lying in the interval [0, 2π] is ______.
