Sin theta + sin 2 theta + sin 3 theta equal to zero
Answers
Answer:
Explanation:
sin θ + sin 2θ + sin 3θ = 0
sin θ + 2sinθcosθ+ (3sinθ-4sin^3θ) =0
sin θ + 2sinθcosθ + sinθ( 3-4sin^2θ)=0
sinθ( 1+2cosθ+ (3-4sin^2θ)=0
sinθ(1+2cosθ +3 -4(1-cos^2θ)=0
sinθ(4-4+2cosθ+4cos^2θ)=0
sinθ(2cosθ+4cos^2θ)=0
sinθ= 0
sinθ = sin 0
θ= nπ
4cos^2θ+2cosθ+0=0
According to quadratic formula,
D= b^2-4ac
D= (2)^2-4(4)(0)
D= 4
cosθ= (-2±2)/8
cos θ= 0 or -1/2
If, cosθ= 0,
Cosθ= cos π/2
θ= (2n+1)π/2
If, cosθ= -1/2
Cosθ= Cos(π-π/3)
Cosθ= 2nπ± Cos(2π/3)
θ= 6nπ/3±2π/3.
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Therefore the solution for sin θ + sin 2θ + sin 3θ = 0 is θ= nπ or θ = nπ + π/2 or θ = 2nπ ± 2π/3 Where n = 0, ±1, ±2, ±3.....
Given:
sin θ + sin 2θ + sin 3θ = 0
To Find:
General solution of the given trigonometric function, sin θ + sin 2θ + sin 3θ = 0
Solution:
The given question can be solved very easily as shown below.
Given a trigonometric function,
⇒ sin θ + sin 2θ + sin 3θ = 0
[ ∵ sin 2θ = 2 sinθcosθ and sin 3θ = 3sinθ-4sin³θ ]
⇒ sin θ + 2sinθcosθ+ (3sinθ-4sin³θ) =0
⇒ sin θ + 2sinθcosθ + sinθ( 3-4sin²θ)=0
⇒ sinθ( 1+2cosθ+ (3-4sin²θ)=0
⇒ sinθ(1+2cosθ +3 -4(1-cos²θ)=0
⇒ sinθ(4-4+2cosθ+4cos²θ)=0
⇒ sinθ(2cosθ+4cos²θ)=0
⇒ sinθ= 0 ⇒ θ= nπ
Or
⇒ 4cos²θ+2cosθ=0
⇒ cosθ (4cosθ + 2 ) = 0
⇒ cosθ = 0 Or (4cosθ + 2 ) = 0
⇒ θ = nπ + π/2 Or θ = 2nπ ± 2π/3
Where n = 0, ±1, ±2, ±3.....
Therefore the solution for sin θ + sin 2θ + sin 3θ = 0 is θ= nπ or θ = nπ + π/2 or θ = 2nπ ± 2π/3 Where n = 0, ±1, ±2, ±3.....
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