find the genearal solution of the following equation 4csc^2x-7=cotx
since csc^2 = 1+cot^2,
4(1+cot^2x)-7 = cotx
4+4cot^2x-7 = cotx
4cot^2x - cotx - 3 = 0
(4cotx+3)(cotx-1) = 0
Now it's easy, right?
(4cotx+3)(cotx-1) = 0
i reached this point i need to know how to find the angles
oh come on. either
4cotx+3 = 0
cotx = -3/4
or
cotx-1 = 0
cotx = 1
In the first case, x will be in QII or QIV
on the second case, x will be in QI or QIII
To find the general solution of the equation 4csc^2x - 7 = cotx, we can rewrite the equation using trigonometric identities.
First, let's rewrite cotx in terms of sine and cosine:
cotx = cosx/sinx
Next, replace csc^2x with its equivalent expression in terms of sine:
csc^2x = 1/sin^2x
Now, substitute these values back into the original equation:
4(1/sin^2x) - 7 = cosx/sinx
To simplify further, we can multiply through by sin^2x, the common denominator:
4 - 7sin^2x = cosx * sinx
Rearranging the terms, we get:
7sin^2x + cosx * sinx - 4 = 0
Now, let's solve this quadratic equation for sinx. We can reassign sinx to a variable, say t, and solve for t:
7t^2 + t - 4 = 0
Factoring the quadratic equation, we have:
(7t - 4)(t + 1) = 0
Setting each factor equal to zero, we get two possible values for t:
7t - 4 = 0 --> t = 4/7
t + 1 = 0 --> t = -1
Now, recall that t represents sinx, so substitute the values back:
sinx = 4/7, and sinx = -1
To find the general solution, we need to determine the possible values for x based on these values for sine.
For sinx = 4/7:
Using the inverse sine function, we write:
x = arcsin(4/7) + 2nπ, where n is an integer.
For sinx = -1:
Using the inverse sine function, we write:
x = arcsin(-1) + 2nπ, where n is an integer.
Now we have the general solution for x in terms of n:
x = arcsin(4/7) + 2nπ, and x = arcsin(-1) + 2nπ.
Therefore, the general solution for the given equation is:
x = arcsin(4/7) + 2nπ, and x = π + 2nπ, where n is an integer.