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