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Question:

The number of values of α in [0, 2π] for which 2sin3α - sin2α + 7sinα = 2, is :

4

3

6

1

Solution:

2sin3α - sin2α + 7sinα = 2
2sin3α - sin2α + 7sinα - 6 = 0
2(sinα - 1)(sin2α + sinα + 1) - sinα(sinα - 1) = 0
(sinα - 1)(2sin2α + 2sinα + 2 - sinα) = 0
(sinα - 1)(2sin2α + sinα + 2) = 0
sinα = 1 or 2sin2α + sinα + 2 = 0
Value '2' is not possible
So, total 3 solutions
α = π/2, 5π/6, π/6