The solutions in the interval [0, 2] is the θ=0,π/2,π,π3/2.
According to the statement
we have to find that the all solutions of the given statement in the interval [0, 2].
So, For this purpose,
So According to the trigonometry
now lets take the inverse sin of both sides to solve,
sin^-1(sin(2θ))=sin^-1(0)
2θ=sin-1(0)
So you can either plug that into a calculator or just remember (from the unit circle) that the sine is equal to zero at 0 AND at π
2θ=0,π
But we're not done because the sine is periodic which means there will be solutions every trip around the unit circle (2π)
So 2θ=0+N^2π and π+N^2π where N=0,+/-1,+/-2,etc...
But we're looking for values of θ so we have to divide everything by 2
θ=0+Nπ and π/2+Nπ
Now we just have to find the values of N that yield solutions on the interval [0,2π]
N=0 gives 0 and π/2
N=1 gives π and π3/2
So the full solution set is
θ=0,π/2,π,π3/2;
So, The solutions in the interval [0, 2] is the θ=0,π/2,π,π3/2.
Learn more about interval here
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