मराठी

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

Advertisements
Advertisements

प्रश्न

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

बेरीज
Advertisements

उत्तर

We have sin θ + sin 3θ + sin 5θ = 0

or (sin θ + sin 5θ) + sin 3θ = 0

or 2 sin 3θ cos 2θ + sin 3θ = 0   

or sin 3θ (2 cos 2θ + 1) = 0

or sin 3θ = 0 or cos 2θ = `- 1/2`

When sin 3θ = 0, then 3θ = nπ or θ = `("n"pi)/3`

When cos 2θ = `-1/2`

= `cos  (2pi)/3`

Then 2θ = `2"n"pi +- (2pi)/3` or θ = `"n"pi +- pi/3`

Which gives θ = `(3"n" + 1)  pi/3` or θ = `(3"n" - 1)  pi/3`

All these values of θ are contained in θ = `("n"pi)/3` , n ∈ Z.

Hence, the required solution set is given by `{θ : θ = ("n"pi)/3, "n" ∈ "Z"}`

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 3: Trigonometric Functions - Solved Examples [पृष्ठ ४२]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 11
पाठ 3 Trigonometric Functions
Solved Examples | Q 7 | पृष्ठ ४२

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

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


Find the general solution of the equation cos 3x + cos x – cos 2x = 0


Find the general solution of the equation  sin x + sin 3x + sin 5x = 0


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


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


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:
cos (A + B) + cos C = 0


Prove that:

\[\sin\frac{10\pi}{3}\cos\frac{13\pi}{6} + \cos\frac{8\pi}{3}\sin\frac{5\pi}{6} = - 1\]

\[\sqrt{\frac{1 + \cos x}{1 - \cos x}}\] 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 sec x + tan x = k, cos x =


Find the general solution of the following equation:

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

Find the general solution of the following equation:

\[\tan px = \cot qx\]

 


Find the general solution of the following equation:

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

Find the general solution of the following equation:

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

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:

\[\cos 4 x = \cos 2 x\]

Solve the following equation:

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

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


Solve the following equation:
3tanx + cot x = 5 cosec x


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


If secx cos5x + 1 = 0, where \[0 < x \leq \frac{\pi}{2}\], find the value of x.


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


Write the set of values of a for which the equation

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

If \[2 \sin^2 x = 3\cos x\]. where \[0 \leq x \leq 2\pi\], then find the value of x.


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


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


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 5x − sin x = cos 3


Choose the correct alternative:
`(cos 6x + 6 cos 4x + 15cos x + 10)/(cos 5x + 5cs 3x + 10 cos x)` is equal to


If a cosθ + b sinθ = m and a sinθ - b cosθ = n, then show that a2 + b2 = m2 + n2 


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×