Find real θ such that.
Answers
Answered by
4
Answer:
3+2isinθ
=
(1+2isinθ)(1−2isinθ)
(3+2isinθ)(1+2isinθ)
=
1−4i
2
sin
2
θ
3+2isinθ+6isinθ+4i
2
sin
2
θ
=
1+4sin
2
θ
3+8isinθ−4sin
2
θ
=
1+4sin
2
θ
3−4sin
2
θ+8isinθ
8sinθ=0⇒sinθ=0
θ=nπ
Step-by-step explanation:
Answered by
0
Answer:
Step-by-step explanation:
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