Question

What is the solution to the trigonometric inequality 2sin(x)+3> sin^2(x)

Answer 1

Answer:

$-\frac{\pi}{2}+2n\pi<x<\frac{3\pi}{2}+2n\pi$

Step-by-step explanation:

Given:

$2\sin(x)+3> \sin^2(x)$

To Find:

The Solution of the above inequality

Explanation:

$2\sin(x)+3> \sin^2(x).......(i)\\\text{let} \sin (x) = u\\\Rightarrow 2u+3> u^2\\\Rightarrow u^2-2u-3<0\\\Rightarrow u^2-3u+u-3<0\\\Rightarrow u(u-3)+1(u-3)<0\\\Rightarrow (u+1)(u-3)<0\\\Rightarrow -1<u<3\\\text{and since} \sin(x) = u\\\Rightarrow -1<\sin(x)<3$

$\text{as if} \;a<u<b \Rightarrow a<u \;\text{and}\; u<b\\\Rightarrow -1<\sin(x) \; \text{and} \; \sin(x) <3\\-1<\sin(x) \; : -\frac{\pi}{2}+2n\pi<x<\frac{3\pi}{2}+2n\pi\\\text{and}\; \sin(x) <3 : \text{true for all } x \in R\\\text{Now, combining the intervals we get}\\-\frac{\pi}{2}+2n\pi<x<\frac{3\pi}{2}+2n\pi \; \text{and} \; \text{true for all } x \in R\\-\frac{\pi}{2}+2n\pi<x<\frac{3\pi}{2}+2n\pi.$