Answer:
Here’s how you can solve the equation 3sin^2x=-sin x +4 for x in the interval [0,2pi):
First, let’s move all the terms to one side of the equation to get 3sin^2x + sin x - 4 = 0. This is a quadratic equation in terms of sin x, so we can solve for sin x using the quadratic formula: (-b ± sqrt(b^2 - 4ac)) / (2a). Plugging in the values for a, b, and c, we get:
sin x = (-1 ± sqrt(1 + 48)) / 6
Solving this gives us two possible values for sin x: -1 and 4/3. However, since the range of the sine function is [-1, 1], the value 4/3 is not a valid solution. So we only need to consider the case where sin x = -1.
Since we’re looking for solutions in the interval [0, 2pi), there is only one value of x that satisfies this equation: x = 3pi/2.
So the solution to the equation 3sin^2x=-sin x +4 in the interval [0,2pi) is x = 3pi/2.
Explanation: