Solve the trigonometric equation3sin(2𝑡)+4=1to find the exact solution on the interval [−𝜋2,𝜋2]

To solve the trigonometric equation 3sin(2t) + 4 = 1, follow these steps:

1. First, isolate the trigonometric function. To do this, subtract 4 from both sides of the equation to get 3sin(2t) = -3.

2. Next, divide both sides by 3 to get sin(2t) = -1.

3. Now, you need to find the value of 2t. The sine function equals -1 at 3π/2 + 2nπ, where n is an integer. So, 2t = 3π/2 + 2nπ.

4. Divide both sides by 2 to solve for t. This gives t = 3π/4 + nπ.

5. Finally, find the values of t in the interval [-π/2, π/2]. The only value in this interval is t = -π/4.