हिंदी

Solve the Following Equation: 3 Cos 2 X − 2 √ 3 Sin X Cos X − 3 Sin 2 X = 0

Advertisements
Advertisements

प्रश्न

Solve the following equation:

\[3 \cos^2 x - 2\sqrt{3} \sin x \cos x - 3 \sin^2 x = 0\]
योग
Advertisements

उत्तर

\[3 \cos^2 x - 2\sqrt{3} \sin x \cos x - 3 \sin^2 x = 0\]

Now,

\[3 ( \cos^2 x - \sin^2 x) - \sqrt{3} \sin2x = 0\]

\[ \Rightarrow 3 \cos2x - \sqrt{3} \sin2x = 0\]

\[ \Rightarrow \sqrt{3} (\sqrt{3} \cos2x - \sin2x) = 0\]

\[ \Rightarrow (\sqrt{3} \cos2x - \sin2x) = 0\]

\[ \Rightarrow \frac{\sin2x}{\cos2x} = \sqrt{3} \]

\[ \Rightarrow \tan2x = \tan \frac{\pi}{3}\]

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

\[ \Rightarrow x = \frac{n\pi}{2} + \frac{\pi}{6}, n \in Z\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 11: Trigonometric equations - Exercise 11.1 [पृष्ठ २२]

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 11 Trigonometric equations
Exercise 11.1 | Q 3.6 | पृष्ठ २२

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

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


If \[\sin x = \frac{a^2 - b^2}{a^2 + b^2}\], then the values of tan x, sec x and cosec x


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


In a ∆ABC, prove that:

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

In a ∆A, B, C, D be the angles of a cyclic quadrilateral, taken in order, prove that cos(180° − A) + cos (180° + B) + cos (180° + C) − sin (90° + D) = 0


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


If \[\frac{\pi}{2} < x < \frac{3\pi}{2},\text{ then }\sqrt{\frac{1 - \sin x}{1 + \sin x}}\] is equal to

 


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


sin6 A + cos6 A + 3 sin2 A cos2 A =


If sec x + tan x = k, cos x =


Which of the following is incorrect?


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

 

Find the general solution of the following equation:

\[\sin x = \frac{1}{2}\]

Find the general solution of the following equation:

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

Find the general solution of the following equation:

\[\tan px = \cot qx\]

 


Solve the following equation:

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

Solve the following equation:

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

Solve the following equation:

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

Solve the following equation:

\[\sqrt{3} \cos x + \sin x = 1\]


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


Solve the following equation:
 cosx + sin x = cos 2x + sin 2x

 


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

 


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

 

Write the solution set of the equation 

\[\left( 2 \cos x + 1 \right) \left( 4 \cos x + 5 \right) = 0\] 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.


The number of values of ​x in [0, 2π] that satisfy the equation \[\sin^2 x - \cos x = \frac{1}{4}\]


If \[\sqrt{3} \cos x + \sin x = \sqrt{2}\] , then general value of x is


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


If \[\cos x = - \frac{1}{2}\] and 0 < x < 2\pi, then the solutions are


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 theta + sqrt(3) cos theta` = 1


Solve the following equations:
`tan theta + tan (theta + pi/3) + tan (theta + (2pi)/3) = sqrt(3)`


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


Solve the equation sin θ + sin 3θ + sin 5θ = 0


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 5cos2θ + 7sin2θ – 6 = 0


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×