Solve the equation exactly: \$3sinθ + 3 = 2cos^2(θ)\$ , 0 ≤ θ <2π.

The given equation is

\$3sinθ + 3 = 2cos^2(θ)\$ , 0 ≤ θ <2π

\$cos^2(\theta) = 1- sin^2(theta)\$.

Now plug the formula in the given equation then simplify

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

\$3sin\theta + 3 = 2 - 2sin^2(theta)\$

Simplify the equation

\$2sin^2\theta + 3sin\theta +1 = 0\$

This is now a quadratic equation in terms of sin(θ). We can solve it using the quadratic formula

\$sin\theta = (− b ± (\sqrt(b^2 - 4ac)))/ (2a) \$

Here a = 2 , b = 3 c = 1

Now plug the values in the formula and simplify

\$sin\theta = (−3 ± (\sqrt(3^2 - 4 times 2 times1))) / (2 times 2) \$

\$sin\theta =( -3 ± (\sqrt(9 - 8))) / 4\$

\$sin\theta = (-3 ± 1) /4\$

This gives us two solutions for sin⁡(θ):

\$sin\theta = (-3 +1) /4 = - 1/2\$

\$sin\theta = (- 3 - 1) / 4 = - 1\$

When \$sin⁡(θ)=−1/2\$. This happens in the third and fourth quadrants. Using the unit circle or inverse sine function, we find two angles

\$\theta_1 = (7pi)/6\$

\$\theta_2 = (11pi)/6\$

When sin⁡(θ) = − 1. This happens when \$θ =(3pi)/2\$. However, θ must lie in the range 0 ≤ θ < 2π. So

\$\theta_3 = (3pi)/2\$

So, the solutions for θ are \$(7pi)/6 , (11pi)/6 and (3pi)/2\$.

This option includes \$(π)/6\$ and \$(11π)/6\$, which are incorrect. As explained in the solution process, these values fall outside the given range (0 ≤ θ < 2π)

\$(pi)/6 , (11pi)/6, (3pi)/2\$

This is incorrect because the angles are not in the range where cosine is positive. The angles \$ (−7π)/6\$ and \$(−11π)/6\$​ fall in the second and third quadrants where cosine is negative

\$(-7pi)/6 , (-11pi)/6, (3pi)/2\$

This option includes \$(π)/2\$, which is incorrect. While cosine is 0 at \$π/2\$, the equation involves sine and its square. Sine is not - 1 at \$π/2\$, so this value wouldn’t satisfy the equation

\$(pi)/6 , (11pi)/6, (pi)/2\$

This is correct because \$(7π)/6\$​ and \$(11π)/6\$​ are in the correct range where cosine is positive, and \$(3π)/2\$​ is a valid angle within the given interval

\$(7pi)/6 , (11pi)/6, (3pi)/2\$

Option

D

Can you summarize what you’ve understood in the above steps?