General solution
Cos^2 +3sin= -3
cos^2 x + 3sinx = -3
1 - sin^2 x + 3sinx + 3 = 0
sin^2 x - 3sinx - 4 = 0
(sinx -4)(sinx + 1) = 0
sinx = 4 , which is not possible
or
sinx = -1
x = 3π/2
since the period of sinx is 2π
general solution:
x = 3π/2 + 2kπ, where k is an integer.
To find the general solution to the equation cos^2(x) + 3sin(x) = -3, we need to manipulate the equation and solve for the values of x that satisfy it.
First, let's work with the equation by using a trigonometric identity to express cos^2(x) in terms of sin(x). The identity we'll use is cos^2(x) + sin^2(x) = 1:
cos^2(x) + 3sin(x) = -3
1 - sin^2(x) + 3sin(x) = -3
Rearranging the terms, we get:
sin^2(x) - 3sin(x) + 4 = 0
Now we have a quadratic equation in terms of sin(x). We can solve this by factoring or by using the quadratic formula.
By factoring, we can rewrite the equation as:
(sin(x) - 1)(sin(x) - 4) = 0
Setting each factor equal to zero individually:
sin(x) - 1 = 0 --> sin(x) = 1
sin(x) - 4 = 0 --> sin(x) = 4 (This equation has no real solutions since the sine function only ranges between -1 and 1)
For sin(x) = 1, we can find the values of x by considering the unit circle or trigonometric values. The angle whose sine is 1 is π/2 or 90 degrees. Also, sine has a periodicity of 2π or 360 degrees, so we can add 2π*n (where n is an integer) to get all the solutions.
Therefore, the general solution for sin(x) = 1 is:
x = π/2 + 2πn
So, the general solution to the equation cos^2(x) + 3sin(x) = -3 is:
x = π/2 + 2πn, where n is an integer.