To solve this equation, we need to use trigonometric identities. Let's start by rewriting the equation in terms of sine and cosine:
7sin^2x - 22(1 - sin^2x) + 10 = 07sin^2x - 22 + 22sin^2x + 10 = 029sin^2x - 12 = 0
Now we can solve for sin^2x:
29sin^2x - 12 = 029sin^2x = 12sin^2x = 12/29
Taking the square root of both sides:
sinx = ±√(12/29)
Therefore, the solutions for x are:
x = arcsin(±√(12/29)) + 2nπwhere n is an integer.
To solve this equation, we need to use trigonometric identities. Let's start by rewriting the equation in terms of sine and cosine:
7sin^2x - 22(1 - sin^2x) + 10 = 0
7sin^2x - 22 + 22sin^2x + 10 = 0
29sin^2x - 12 = 0
Now we can solve for sin^2x:
29sin^2x - 12 = 0
29sin^2x = 12
sin^2x = 12/29
Taking the square root of both sides:
sinx = ±√(12/29)
Therefore, the solutions for x are:
x = arcsin(±√(12/29)) + 2nπ
where n is an integer.