Q.If $2\sin^2\theta = 3\cos\theta$, where $0 \le \theta \le 2\pi$, then find the value of $\theta$.

The key idea is to solve a trigonometric equation by transforming it into a quadratic equation in terms of a single trigonometric function.

First, use the identity $\sin^2\theta = 1 - \cos^2\theta$ to rewrite the equation entirely in terms of $\cos\theta$:

$2(1 - \cos^2\theta) = 3\cos\theta$

$2 - 2\cos^2\theta = 3\cos\theta$

Rearrange this into a standard quadratic form:

$2\cos^2\theta + 3\cos\theta - 2 = 0$

Let $x = \cos\theta$. The equation becomes $2x^2 + 3x - 2 = 0$. Factor this quadratic equation:

$(2x - 1)(x + 2) = 0$

This gives two possible values for $x$: $x = \frac{1}{2}$ or $x = -2$.

Substitute back $x = \cos\theta$:

$\cos\theta = \frac{1}{2}$ or $\cos\theta = -2$. …

When faced with a trigonometric equation involving different powers or functions of the same angle, a common strategy is to simplify it into an equation involving a single trigonometric function. This often allows us to transform the problem into a more familiar algebraic form, such as a quadratic equation.

In this problem, we have $\sin^2\theta$ and $\cos\theta$. The fundamental trigonometric identity $\sin^2\theta + \cos^2\theta = 1$ is key here. By using this identity, we can express $\sin^2\theta$ in terms of $\cos^2\theta$, thereby converting the entire equation into one solely involving $\cos\theta$. Once we have an equation in terms of a single trigonometric function, we can treat that function (e.g., $\cos\theta$) as a variable and solve the resulting algebraic equation. Finally, we find the angles $\theta$ that satisfy these trigonometric values within the specified domain.

Here is the step-by-step solution:

1. Convert the equation to a single trigonometric function.

   The given equation is $2\sin^2\theta = 3\cos\theta$.

   We know the Pythagorean identity:

   $\sin^2\theta + \cos^2\theta = 1$

   From this, we can write $\sin^2\theta = 1 - \cos^2\theta$.

   Substitute this into the original equation:

   $2(1 - \cos^2\theta) = 3\cos\theta$

2. Rearrange into a quadratic equation.

   Expand the left side and move all terms to one side to form a standard quadratic equation:

   $2 - 2\cos^2\theta = 3\cos\theta$

   $0 = 2\cos^2\theta + 3\cos\theta - 2$

   This is a quadratic equation in terms of $\cos\theta$.

3. Solve the quadratic equation.

   Let $x = \cos\theta$. The equation becomes:

   $2x^2 + 3x - 2 = 0$

   We can solve this quadratic equation by factoring. We look for two numbers that multiply to $2 \times (-2) = -4$ and add up to $3$. These numbers are $4$ and $-1$.

   $2x^2 + 4x - x - 2 = 0$

   Factor by grouping:

   $2x(x + 2) - 1(x + 2) = 0$

   $(2x - 1)(x + 2) = 0$

   This gives two possible solutions for $x$:

   $2x - 1 = 0 \implies x = \frac{1}{2}$

   $x + 2 = 0 \implies x = -2$ …