मराठी

Find the General Solution of the Following Equation: Sin 2 X = Cos 3 X

Advertisements
Advertisements

प्रश्न

Find the general solution of the following equation:

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

उत्तर

We have:

\[\sin2x = \cos3x\]
\[\cos3x = \sin2x\]

⇒ \[\cos3x = \cos \left( \frac{\pi}{2} - 2x \right)\]

⇒ \[3x = 2n\pi \pm \left( \frac{\pi}{2} - 2x \right), n \in Z\]

On taking positive sign, we have:
\[3x = 2n\pi + \left( \frac{\pi}{2} - 2x \right)\]

⇒ \[5x = 2n\pi + \frac{\pi}{2}\]

⇒ \[x = \frac{2n\pi}{5} + \frac{\pi}{10}\]

⇒ \[x = (4n + 1)\frac{\pi}{10}\]

\[n \in Z\]

 Now, on taking negative sign, we have:

\[3x = 2n\pi - \frac{\pi}{2} + 2x, n \in Z\]
⇒ \[x = 2n\pi - \frac{\pi}{2}\]
⇒ \[x = (4n - 1)\frac{\pi}{2}, n \in Z\]
shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 11: Trigonometric equations - Exercise 11.1 [पृष्ठ २१]

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 11
पाठ 11 Trigonometric equations
Exercise 11.1 | Q 2.04 | पृष्ठ २१

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

If \[x = \frac{2 \sin x}{1 + \cos x + \sin x}\], then prove that

\[\frac{1 - \cos x + \sin x}{1 + \sin x}\] is also equal to a.

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 \[\sin x + \cos x = m\], then prove that \[\sin^6 x + \cos^6 x = \frac{4 - 3 \left( m^2 - 1 \right)^2}{4}\], where \[m^2 \leq 2\]


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: \[\tan\frac{11\pi}{3} - 2\sin\frac{4\pi}{6} - \frac{3}{4} {cosec}^2 \frac{\pi}{4} + 4 \cos^2 \frac{17\pi}{6} = \frac{3 - 4\sqrt{3}}{2}\]

 


Prove that:

\[3\sin\frac{\pi}{6}\sec\frac{\pi}{3} - 4\sin\frac{5\pi}{6}\cot\frac{\pi}{4} = 1\]

 


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

 

Prove that:
\[\sec\left( \frac{3\pi}{2} - x \right)\sec\left( x - \frac{5\pi}{2} \right) + \tan\left( \frac{5\pi}{2} + x \right)\tan\left( x - \frac{3\pi}{2} \right) = - 1 .\]


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


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 =

 


Find the general solution of the following equation:

\[\tan x = - \frac{1}{\sqrt{3}}\]

Find the general solution of the following equation:

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

Find the general solution of the following equation:

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

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:

\[2 \sin^2 x + \sqrt{3} \cos x + 1 = 0\]

Solve the following equation:

\[\tan^2 x + \left( 1 - \sqrt{3} \right) \tan x - \sqrt{3} = 0\]

Solve the following equation:

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

Solve the following equation:

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

Solve the following equation:

\[\sin 3x - \sin x = 4 \cos^2 x - 2\]

Solve the following equation:

\[\sin 2x - \sin 4x + \sin 6x = 0\]

Solve the following equation:
\[5 \cos^2 x + 7 \sin^2 x - 6 = 0\]


Solve the following equation:
3 – 2 cos x – 4 sin x – cos 2x + sin 2x = 0


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


Write the general solutions of tan2 2x = 1.

 

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

 

If \[3\tan\left( x - 15^\circ \right) = \tan\left( x + 15^\circ \right)\] \[0 < x < 90^\circ\], find θ.


If \[\tan px - \tan qx = 0\], then the values of θ form a series in

 


General solution of \[\tan 5 x = \cot 2 x\] is


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


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


Solve the following equations:
cos θ + cos 3θ = 2 cos 2θ


Choose the correct alternative:
If sin α + cos α = b, then sin 2α is equal to


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×