solve the equation
3 sin^2(theta) -sin(theta)-4=0
To solve the equation 3sin²(theta) - sin(theta) - 4 = 0, we can use a substitution to make the equation easier to solve. Let's substitute u = sin(theta), so the equation becomes:
3u² - u - 4 = 0
Now we have a quadratic equation in terms of u. We can solve this equation by factoring, completing the square, or using the quadratic formula.
To factor the equation, we need to find two numbers whose product is -12 (3 * -4) and whose sum is -1 (the coefficient of the middle term). The numbers -4 and 3 satisfy these conditions:
3u² - 4u + 3u - 4 = 0
Now we can factor by grouping:
(3u² - 4u) + (3u - 4) = 0
u(3u - 4) + 1(3u - 4) = 0
(3u - 4)(u + 1) = 0
Now we have two separate equations:
3u - 4 = 0 or u + 1 = 0
Solving each equation:
3u - 4 = 0
3u = 4
u = 4/3
u + 1 = 0
u = -1
Now, we substitute back u = sin(theta):
sin(theta) = 4/3 or sin(theta) = -1
However, the range of sine function is between -1 and 1, so sin(theta) cannot be 4/3. Therefore, there is no solution for the equation sin(theta) = 4/3.
For sin(theta) = -1, we can find the values of theta that satisfy the equation. Since sin(theta) = -1, theta must be in the form (2n + 1)π, where n is an integer.
Therefore, the solutions for the equation 3sin²(theta) - sin(theta) - 4 = 0 are:
theta = (2n + 1)π, where n is an integer.