English

Find the general solution of sin θ + sin 3θ + sin 5θ = 0

Advertisements
Advertisements

Question

Find the general solution of sin θ + sin 3θ + sin 5θ = 0

Sum
Advertisements

Solution

We have sin θ + sin 3θ + sin 5θ = 0

∴ (sin θ + sin 5θ) + sin 3θ = 0

∴ 2 sin 3θ × cos 2θ + sin 3θ = 0

∴ (2 cos 2θ + 1) sin 3θ = 0

∴ sin 3θ = 0 or cos 2θ = `-1/2`

∴ sin 3θ = 0 or cos 2θ = `cos  (2π)/3`

∴ 3θ = nπ or 2θ = `2nπ ± (2π)/3`, where n ∈ Z.

θ = `(nπ)/3` or θ = `nπ ± π/3`,

where n ∈ Z is the required general solution.

shaalaa.com
  Is there an error in this question or solution?
2022-2023 (March) Official

RELATED QUESTIONS

Find the principal solution of the following equation: 

cosθ = `(1)/(2)`


Find the principal solution of the following equation:

cot θ = 0


Find the principal solution of the following equation:

sin θ = `-1/2`


Find the general solution of the following equation:

tan θ = `(1)/(sqrt(3))`


Find the general solution of the following equation:

cot θ = 0.


Find the general solution of the following equation:

tan θ = - 1


Find the general solution of the following equation:

4sin2θ = 1.


Find the general solution of the following equation:

cosθ + sinθ = 1.


State whether the following equation have solution or not?

cos 2θ = – 1


State whether the following equation has a solution or not?

cos2θ = – 1.


State whether the following equation have solution or not?

3 tanθ = 5


With the usual notations prove that `2{asin^2  "C"/(2) + "c"sin^2  "A"/(2)}` = a – b + c.


In ΔABC, if a cos A = b cos B then prove that the triangle is either a right angled or an isosceles traingle.


With usual notations prove that 2(bc cosA + ac cosB + ab cosC) = a2 + b2 + c.


Select the correct option from the given alternatives:

The principal solutions of equation cot θ = `sqrt3` are ______.


Select the correct option from the given alternatives:

If polar coordinates of a point are `(2, pi/4)`, then its cartesian coordinates are


Select the correct option from the given alternatives:

`"tan"(2"tan"^-1 (1/5) - pi/4)` = ______


The principal value branch of sec-1x is ______.


Select the correct option from the given alternatives:

In any ΔABC, if acos B = bcos A, then the triangle is


Find the general solutions of the following equation:

`tan^2 theta = 3`


In ΔABC, prove that `("a - b")^2 cos^2  "C"/2 + ("a + b")^2 sin^2  "C"/2 = "c"^2`


If 2 tan-1(cos x) = tan-1(2 cosec x), then find the value of x.


Show that `tan^-1  1/2 - tan^-1  1/4 = tan^-1  2/9`.


If | x | < 1, then prove that

`2 tan^-1 "x" = tan^-1 ("2x"/(1 - "x"^2)) = sin^-1 ("2x"/(1 + "x"^2)) = cos^-1 ((1 - "x"^2)/(1 + "x"^2))`


If x, y, z are positive, then prove that

`tan^-1 (("x - y")/(1 + "xy")) + tan^-1 (("y - z")/(1 + "yz")) + tan^-1 (("z - x")/(1 + "zx")) = 0`


Find the principal solutions of sin x − 1 = 0


Find the principal solutions of cos 2𝑥 = 1


Find the principal solutions of sin x = `-1/2`


Find the principal solutions of tan x = `-sqrt(3)`


If 2 cos2 θ + 3 cos θ = 2, then permissible value of cos θ is ________.


The number of solutions of `sin^2 theta = 1/2` in [0, π] is ______.


The number of solutions of cos 2θ = sin θ in (0, 2π) is ______


The general solution of sin 2x = cos 2x is ______ 


The general solution of the equation tan θ tan 2θ = 1 is given by ______ 


Which of the following is true in a triangle ABC?


If sin θ + cos θ = 1, then the general value of θ is ______.


The general solution of cot θ + tan θ = 2 is ______.


The equation 3sin2x + 10 cos x – 6 = 0 is satisfied, if ______.


If 2 tan–1(cos x) = tan–1(2 cosec x). then find the value of x.


The general solution of the equation tan2 x = 1 is ______.


The general solution of sin x – 3 sin 2x + sin 3x = cos x – 3 cos 2x + cos 3x is ______.


Prove that the general solution of cos θ = cos α is θ = 2nπ ± α, n ∈ Z.


The general solution of cot 4x = –1 is ______.


If `tanx/(tan 2x) + (tan 2x)/tanx + 2` = 0, then the general value of x is ______.


If `sin^-1  4/5 + cos^-1  12/13` = sin–1α, then find the value of α.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×