Trigonometric equation $2\sin^3(x) + 3 = \cos^2(x) + 5\sin(x)$

Question

I'm having trouble solving this equation:

$$ 2\sin^3(x) + 3 = \cos^2(x) + 5\sin(x) $$

Some hint would be appreciated :) Thanks in advance!

Answer 1

Hint : Use $$\cos ^{ 2 }{ x } =1-\sin ^{ 2 }{ x } \\ 3=3\cdot 1=3\left( \cos ^{ 2 }{ x } +\sin ^{ 2 }{ x } \right) $$

Answer 2

Hint:

Use the identity $\cos^2(x)\equiv 1-\sin^2(x)$:

$$2\sin^3(x)+3=1-\sin^2(x)+5\sin(x)$$

Now, you can substitute $u=\sin(x)$ to obtain a cubic equation:

$$2u^3+3=1-u^2+5u$$ $$2u^3+u^2-5u+2=0$$

Now, you can solve the cubic equation.

Answer 3

Since $$ \cos^2x=1-\sin^2x, $$ the equation is equivalent to: \begin{eqnarray} 2\sin^3x+3&=&1-\sin^2x+5\sin x\\ 2\sin^3x+\sin^2x-5\sin x+2&=&0\\ 2\sin^3x+\sin^2x-3\sin x-2\sin x+2&=&0\\ (2\sin^2x+\sin x-3)\sin x-2(\sin x-1)&=&0\\ (2\sin^2x-2+\sin x-1)\sin x-2(\sin x-1)&=&0\\ [2(\sin^2x-1)+\sin x-1]\sin x-2(\sin x-1)&=&0\\ [2(\sin x+1)(\sin x-1)+\sin x-1]\sin x-2(\sin x-1)&=&0\\ (\sin x-1)(2\sin x+2+1)\sin x-2(\sin x-1)&=&0\\ (\sin x-1)[(2\sin x+3)\sin x-2]&=&0\\ (\sin x-1)(2\sin^2x+3\sin x-2)&=&0\\ \sin x-1&=&0 \text{ or } 2\sin^2x+3\sin x-2=0\\ \sin x&=&1 \text{ or } \sin x=\dfrac{-3\pm5}{4}\\ \sin x&=&1 \text{ or } \sin x=\dfrac{1}{2} \text{ or } \sin x=-2\\ \sin x&=&1 \text{ or } \sin x=\dfrac{1}{2} \text{ because } \sin x \in [-1,1] \end{eqnarray} Solving the two equations we get $$ x=\dfrac{\pi}{2}+2k\pi \text{ or } x=\dfrac{\pi}{6}+2k\pi \text{ or } x=\dfrac{5\pi}{6}+2k\pi \quad k \in \mathbb{Z}. $$