Solve for: 2cos(x)^2+3sin(x)-3=0

Expression: $2{\cos\left({x}\right)}^{2}+3\sin\left({x}\right)-3=0$

Use ${\cos\left({t}\right)}^{2}=1-{\sin\left({t}\right)}^{2}$ to expand the expression

$2\left( 1-{\sin\left({x}\right)}^{2} \right)+3\sin\left({x}\right)-3=0$

Distribute $2$ through the parentheses

$2-2{\sin\left({x}\right)}^{2}+3\sin\left({x}\right)-3=0$

Calculate the difference

$-1-2{\sin\left({x}\right)}^{2}+3\sin\left({x}\right)=0$

To get an equation that is easier to solve, substitute $t$ for $\sin\left({x}\right)$

$-1-2{t}^{2}+3t=0$

Solve the equation for $t$

$\begin{array} { l }t=\frac{ 1 }{ 2 },\\t=1\end{array}$

Substitute back $t=\sin\left({x}\right)$

$\begin{array} { l }\sin\left({x}\right)=\frac{ 1 }{ 2 },\\\sin\left({x}\right)=1\end{array}$

Solve the equation for $x$

$\begin{array} { l }\begin{array} { l }x=\frac{ π }{ 6 }+2kπ,& k \in ℤ\end{array},\\\begin{array} { l }x=\frac{ 5π }{ 6 }+2kπ,& k \in ℤ\end{array},\\\sin\left({x}\right)=1\end{array}$

Solve the equation for $x$

$\begin{array} { l }\begin{array} { l }x=\frac{ π }{ 6 }+2kπ,& k \in ℤ\end{array},\\\begin{array} { l }x=\frac{ 5π }{ 6 }+2kπ,& k \in ℤ\end{array},\\\begin{array} { l }x=\frac{ π }{ 2 }+2kπ,& k \in ℤ\end{array}\end{array}$