7sin2x-22cos^2x+10=0

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вопрос

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 =
7sin^2x - 22 + 22sin^2x + 10 =
29sin^2x - 12 = 0

Now we can solve for sin^2x:

29sin^2x - 12 =
29sin^2x = 1
sin^2x = 12/29

Taking the square root of both sides:

sinx = ±√(12/29)

Therefore, the solutions for x are:

x = arcsin(±√(12/29)) + 2n
where n is an integer.