what are all the exact solutions of 2 sin^2 x- sin x = 0
Question
Answer:
Factor:2sin^2x-sinx = 0
sinx(2sinx - 1) = 0
Therefore the solutions are when:
sin x = 0
And
sinx = 1/2
So sinx = 0
is true when x = 0 and pi and all the angles coterminal with these points. Thus, the answer is x = pi*n, (where n is some integer)
sinx = 1/2
is true when x = pi/6 and 5pi/6 and the angles coterminal with these points.
Thus, the answer is x = pi/6 + 2pi*n (where n is some integer)
and x = 5pi/6 + 2pi*n (where n is some integer)
solved
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10 months ago
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