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 $\left(2\cos \left(x\right)-1\right)\left(\cos \left(x\right)-1\right)\=0$.

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≻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≻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.