Find all solution of the equation in the interval [0,2\pi]. 2\sin3

Tabansi

Tabansi

Answered question

2021-08-17

Find all solution of the equation in the interval [0,2π].
2sin3θ+2=0

Answer & Explanation

gotovub

gotovub

Skilled2021-08-18Added 98 answers

Given:
2sin3θ+2=0[0,2π]
Subtract 2 both the sides
2sin3θ+22=02
2sin3θ=2
Divide 2 both the sides
sin3θ=22
General solution for sin3θ=22
3θ=5π4+2πn,3θ=7π4+2nπ
Solve 3θ=5π4+2πn
θ=5π12+2πn3
and
3θ=7π4+2πn
θ=7π12+2πn3
Solution
θ=5π12,θ=7π12,θ=13π12,θ=5π4,θ=7π4,θ=23π12
Jeffrey Jordon

Jeffrey Jordon

Expert2022-01-31Added 2605 answers

Answer is given below (on video)

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