# Solve the inequation: 4sin^2 x - 8sin x - 5 > 0x ∈ [0, 2π]

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The inequation 4sin^2 x - 8sin x - 5 > 0 has to be solved.

4sin^2 x - 8sin x - 5 > 0

=> 4sin^2 x - 10*sin x + 2sin x - 5 > 0

=> 2*sin x( 2*sin x - 5) + 1(2*sin x - 5) > 0

=> (2*sin x + 1)( 2*sin x - 5) > 0

For this to be true either

(2*sin x + 1) > 0 and ( 2*sin x - 5) > 0

=> sin x > -1/2 and sin x > 5/2

=> sin x > 5/2

which is not possible

or

(2*sin x + 1) < 0 and ( 2*sin x - 5) < 0

sin x < -1/2 and sin x < 5/2

As the minimum value of sin x can be -1, sin x lies in [-1, -1/2)

x lies in (7*pi/6, 11*pi/6)

**The solution of the inequation is (7*pi/6, 11*pi/6)**

It was supposed to be 4sin^2 x - 8sin x - 5 > 0