22 mark for review the function f is given by f(θ)=2cos²θ + 3sinθ - 3. for what values of θ on 0 ≤ θ ≤ 2π…

Explanation:

Step1: Use the identity $\cos^{2}\theta=1 - \sin^{2}\theta$

Substitute $\cos^{2}\theta$ in $f(\theta)$: $f(\theta)=2(1 - \sin^{2}\theta)+3\sin\theta - 3=0$. Expand to get $2-2\sin^{2}\theta+3\sin\theta - 3 = 0$. Rearrange to $2\sin^{2}\theta-3\sin\theta + 1=0$.

Step2: Let $x = \sin\theta$

The equation becomes $2x^{2}-3x + 1=0$. Factor the quadratic equation: $(2x - 1)(x - 1)=0$.

Step3: Solve for $x$

Set each factor equal to zero: For $2x - 1=0$, we have $x=\frac{1}{2}$. For $x - 1=0$, we have $x = 1$.

Step4: Substitute back $\sin\theta$ for $x$

When $\sin\theta=\frac{1}{2}$, $\theta=\frac{\pi}{6}$ or $\theta=\frac{5\pi}{6}$ in the interval $0\leq\theta\leq2\pi$. When $\sin\theta = 1$, $\theta=\frac{\pi}{2}$ in the interval $0\leq\theta\leq2\pi$.

Answer:

C. $\frac{\pi}{6},\frac{\pi}{2}$, and $\frac{5\pi}{6}$