मराठी

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

Advertisements
Advertisements

प्रश्न

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

बेरीज
Advertisements

उत्तर

Given that: sinx – 3sin2x + sin3x = cosx – 3cos2x + cos3x

⇒ (sin3x + sinx) – 3sin2x = (cos3x + cosx) – 3cos2x

⇒ `2sin((3x + x)/2) . cos((3x - x)/2) - 3sin2x = 2cos((3x + x)/2).cos((3x - x)/2) - 3cos2x`

⇒ 2sin2x . cosx – 3sin2x = 2cos2x . cosx – 3cos2x

⇒ 2sin2x cosx – 2cos2x . cosx = 3sin2x – 3cos2x

⇒ 2cosx (sin2x – cos2x) = 3(sin2x – cos2x)

⇒ 2cosx(sin2x – cos2x) – 3(sin2x – cos2x) = 0

⇒ (sin2x – cos2x)(2cosx – 3) = 0

⇒ sin2x – cos2x = 0 and 2cosx – 3 ≠ 0   ....[∵ – 1 ≤ cos x ≤ 1]

⇒ `(sin2x)/(cos2x) - 1` = 0

⇒ tan2x = 1

⇒ tan2x = `tan  pi/4`

⇒ 2x = `npi + pi/4`

∴ x = `(npi)/2 + pi/8`

Hence, the general solution of the equation is x = `(npi)/2 + pi/8`, n ∈ Z.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 3: Trigonometric Functions - Exercise [पृष्ठ ५५]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 11
पाठ 3 Trigonometric Functions
Exercise | Q 28 | पृष्ठ ५५

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

Find the principal and general solutions of the equation sec x = 2


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


If \[\tan x = \frac{a}{b},\] 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 \[\sin x + \cos x = m\], 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 that:  tan 225° cot 405° + tan 765° cot 675° = 0


Prove that: cos 24° + cos 55° + cos 125° + cos 204° + cos 300° = \[\frac{1}{2}\]


In a ∆ABC, prove that:

\[\cos\left( \frac{A + B}{2} \right) = \sin\frac{C}{2}\]

 


\[\sqrt{\frac{1 + \cos x}{1 - \cos x}}\] 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 \[cosec x - \cot x = \frac{1}{2}, 0 < x < \frac{\pi}{2},\]

 

If sec x + tan x = k, cos x =


Find the general solution of the following equation:

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

Find the general solution of the following equation:

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

Find the general solution of the following equation:

\[\tan 2x \tan x = 1\]

Find the general solution of the following equation:

\[\tan mx + \cot nx = 0\]

Solve the following equation:

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

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


Write the general solutions of tan2 2x = 1.

 

Write the set of values of a for which the equation

\[\sqrt{3} \sin x - \cos x = a\] has no solution.

Write the number of points of intersection of the curves

\[2y = - 1 \text{ and }y = cosec x\]

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].

The smallest positive angle which satisfies the equation ​

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

If \[4 \sin^2 x = 1\], then the values of x are

 


If \[e^{\sin x} - e^{- \sin x} - 4 = 0\], then x =


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°

sin4x = sin2x


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


Solve the following equations:
cot θ + cosec θ = `sqrt(3)`


Solve the following equations:
cos 2θ = `(sqrt(5) + 1)/4`


Solve the following equations:
2cos 2x – 7 cos x + 3 = 0


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


In a triangle ABC with ∠C = 90° the equation whose roots are tan A and tan B is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×