Curve tracing

Geometry Level 3

( 1 − 1 2 sin ⁡ x ) ( cos ⁡ 2 x ) 2 = 2 sin ⁡ x − 3 + 1 sin ⁡ x \left(1-\frac{1}{2\sin x}\right)(\cos 2x)^{2}=2\sin x-3+\frac{1}{\sin x}

Find the number of solutions of the equation above in the interval [ 0 , 4 π ] [0,4\pi] .

5 0 More than 5 4 3

Shivam Jadhav by Shivam Jadhav , 22, India

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1 solution

1 − 1 2 s i n x ( c o s 2 x ) 2 = 2 ( s i n x ) 2 − 3 s i n x + 1 s i n x 1-\frac{1}{2sinx}(cos2x)^{2}=2(sinx)^{2}-3sinx+\frac{1}{sinx}
simplifies to:
( c o s 2 x ) 2 = 2 ( s i n x − 1 ) (cos2x)^{2}=2(sinx-1) which can be further simplified to: 2 ( s i n x ) 4 − 2 ( s i n x ) 2 − s i n x + 1 = 0 2(sinx)^{4}-2(sinx)^{2}-sinx+1=0 we can see that this is biquadratic equation whose critical points are all in the range -1 to 1 and the function changes sign 4 times therefore has 4 solutions.

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