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Geometry Level 3

The number of solutions are in [0,2π] such that (sin(2x))^4 = 1/8 is


The answer is 8.

Parth Lohomi by Parth Lohomi , 20, India

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1 solution

s i n 4 ( 2 x ) − 1 8 = 0 sin^{4}(2x)-\frac{1}{8}=0 = > ( s i n 2 ( 2 x ) + 1 8 ) ( s i n ( 2 x ) + 1 8 4 ) ( s i n ( 2 x ) − 1 8 4 ) = 0 => (sin^{2}(2x)+\sqrt{\frac{1}{8}})(sin(2x)+\sqrt[4]{\frac{1}{8}})(sin(2x)-\sqrt[4]{\frac{1}{8}})=0 . If we draw the graph of s i n ( 2 x ) sin(2x) until 2 π 2\pi it has four intersections with y = 1 8 4 y=\sqrt[4]{\frac{1}{8}} and other four with y = − 1 8 4 y=-\sqrt[4]{\frac{1}{8}}

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