Question

solve 2sin^2x=1 over domain of -4pi (greater than or equal to) x (greater than or equal to) -3pi


give exact answers.. i am a visual learner.. if u can draw out the pi chart that would be great.. domain always messes me thanks

Answer 1

\[2\sin^2x=1\\
\sin^2x={1\over2}\\
\implies \sin x=\pm{1\over\sqrt{2}}\]
now you can get the "x"