sin(2) - cos() = 0
=> 2sin()cos() - cos() = 0
=> cos() (2sin() - 1) = 0
=> cos() = 0 or (2sin() - 1) = 0
cos() = 0
We know that cos x = 0 for x = /2 and 3/2
Therefore, solutions of for which cos() = 0 is given by
= /2, = 3/2
(2sin() - 1) = 0
=> sin = 1/2
We know that sine function is positive in the first and second quadrant and sin x = 1/2 for x = /6
Reference angle in the first quadrant = x
Reference angle in the second quadrant = - x
Therefore, solutions of for which (2sin() - 1) = 0 is given by
= /6
and, = - /6 = 5/6
Combining all the solutions of we get
= /6, /2, 5/6, 3/2
Note : The solution for the equation is to be provided in the interval [0, 2)
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