The given trigonometric equation is:

$2\sin^2(x) - 3\sin(x) + 1 = 0$

We need to solve this equation for $x$ in the interval $[0, 2\pi)$.

Step 1: Let $u = \sin(x)$

This transforms the equation into a quadratic form: $2u^2 - 3u + 1 = 0$

Step 2: Solve the quadratic equation

We solve this quadratic equation using the quadratic formula: $u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ For the equation $2u^2 - 3u + 1 = 0$, the coefficients are $a = 2$, $b = -3$, and $c = 1$.

Substitute these values into the quadratic formula: $u = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(1)}}{2(2)}$ $u = \frac{3 \pm \sqrt{9 - 8}}{4}$ $u = \frac{3 \pm \sqrt{1}}{4}$ $u = \frac{3 \pm 1}{4}$

Thus, the two possible values for $u$ (which represent $\sin(x)$) are: $u_1 = \frac{3 + 1}{4} = 1 \quad \text{and} \quad u_2 = \frac{3 - 1}{4} = \frac{1}{2}$

Step 3: Solve for $x$

• For $u_1 = 1$, we solve $\sin(x) = 1$: $x = \frac{\pi}{2}$
• For $u_2 = \frac{1}{2}$, we solve $\sin(x) = \frac{1}{2}$: $x = \frac{\pi}{6}, \, \frac{5\pi}{6}$ (since $\sin(x) = \frac{1}{2}$ at these angles in the interval $[0, 2\pi)$).

Final Answer:

The solutions for $x$ are: $x = \frac{\pi}{2}, \frac{\pi}{6}, \frac{5\pi}{6}$

Would you like more details or have any further questions?

Related Questions:

1. How do you apply the quadratic formula to solve trigonometric equations?
2. Why are there two solutions for $\sin(x) = \frac{1}{2}$?
3. What are the general solutions for $\sin(x) = 1$ and $\sin(x) = \frac{1}{2}$?
4. How do we determine solutions in the interval $[0, 2\pi)$?
5. What are some other methods for solving trigonometric equations?

Tip:

When solving trigonometric equations, it helps to first transform the equation into a quadratic or factorizable form if possible. This simplifies finding the roots.