Questions: Solve the following equation for all solutions in the interval [0,2 π) 4 sin ^2(x)-3=0 π/3, 2 π/3, 4 π/3, 5 π/3 2 π/3, 4 π/3 π/3 π/3, 5 π/3 Solve the following equation for all solutions (not just ones on the interval [0,2 π )) 2 sin (2 x)-√3=0 π/3+2 π n, 5 π/6+2 π n π/6+2 π n, 5 π/6+2 π n π/3+π n, 5 π/3+π n π/6+π n, π/3+π n

Solution Steps

To solve the given trigonometric equations, we will follow these steps:

1. For the first equation $4 \sin^2(x) - 3 = 0$:

   • Isolate $\sin^2(x)$ and solve for $\sin(x)$.
   • Find the values of $x$ in the interval $[0, 2\pi)$ that satisfy the equation.

2. For the second equation $2 \sin(2x) - \sqrt{3} = 0$:

   • Isolate $\sin(2x)$ and solve for $2x$.
   • Find the general solutions for $x$ by considering the periodicity of the sine function.

We start with the equation:

$$4 \sin^2(x) - 3 = 0$$

Isolating $\sin^2(x)$:

$$\sin^2(x) = \frac{3}{4}$$

Taking the square root gives:

$$\sin(x) = \pm \frac{\sqrt{3}}{2}$$

The solutions for $x$ in the interval $[0, 2\pi)$ are:

$$x = \frac{\pi}{3}, \quad x = \frac{2\pi}{3}, \quad x = \frac{4\pi}{3}, \quad x = \frac{5\pi}{3}$$

Thus, the solutions are:

$$\left\{ \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3} \right\}$$

Next, we solve the equation:

$$2 \sin(2x) - \sqrt{3} = 0$$

Isolating $\sin(2x)$:

$$\sin(2x) = \frac{\sqrt{3}}{2}$$

The general solutions for $2x$ are:

$$2x = \frac{\pi}{3} + 2\pi n \quad \text{and} \quad 2x = \frac{2\pi}{3} + 2\pi n$$

Dividing by 2 gives the solutions for $x$:

$$x = \frac{\pi}{6} + \pi n \quad \text{and} \quad x = \frac{\pi}{3} + \pi n$$

Thus, the general solutions are:

$$\left\{ \frac{\pi}{6} + \pi n, \frac{\pi}{3} + \pi n \right\}$$

Final Answer