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explain what is wrong with the following solution for the equation sin2theta=a in the interval {0,2pi} sin 2 theta =1 sin theta = 1/2 theta = pi/6 or theta = 5pi/6 helllpppp em plz
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\(sin(2\theta) =1\rightarrow sin \theta =\dfrac{1}{2}\) is wrong
it should be sin (\(2\theta)\) = 1 \(\rightarrow 2\theta = arcsin (1)= \dfrac{\pi}{2}\) so, \(\theta = \dfrac{\pi}{4}\)
other solution is 3pi/4
thanks can u help me with one more plz
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