Figure A-7.4(a), a graph of $f\left(x\right)\=2 {\sin }^2\left(x\right)\+3 \cos \left(x\right)-3$, suggests that the given equation has four solutions in the interval $\left[0\,2 \mathrm{\π}\right]$, and that $x\=0$ and $x\=2 \mathrm{\π}$ might well be two of these four solutions.

However, two different functions appear in the equation, and the trig identity ${\sin }^2\left(x\right)\=1-{\cos }^2\left(x\right)$ must be used to convert the equation to one quadratic in $\cos \left(x\right)$.

The transformed equation is then $2 {\cos }^2\left(x\right)-3 \cos \left(x\right)\+1\=0$, which factors to $\left(2\cos \left(x\right)-1\right)\left(\cos \left(x\right)-1\right)\=0$.

By the Zero Principle, either or both factors must themselves be zero, so the following two equations must be solved.

The solutions of the first are

The solutions of the second are

Tools≻Load Package: Student Calculus 1

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

Expression palette: Evaluation template
Evaluate equation $$ with ${\sin }^2\left(x\right)\=1-{\cos }^2\left(x\right)$.
Press the Enter key.

Context Panel: Move to Right

Context Panel: Right-hand Side

Context Panel: Factor

Control-drag the first factor equate to zero.
Press the Enter key.

Context Panel:
Student Calculus 1≻Solve≻Find Roots
Enter bounds as per Figure A-7.4(b)

Context Panel: Approximate≻5 (digits)

Control-drag the second factor; equate to zero.
Press the Enter key.

Context Panel:
Student Calculus 1≻Solve≻Find Roots
Enter bounds as per Figure A-7.4(b)

Tools≻Load Package: Student Calculus 1
(Skip this step if package already loaded.)

Assign the equation to the name $\mathrm{q\_\_1}$.

Apply the Roots command with the appropriate bound on $x$.

Apply the evalf command with 5 as the optional digits parameter.

Apply the factor command to the equation.

Use the lhs (left-hand side) command to select the left side of the factored equation.