3 cos в квадрате x = 7(sinx+1)

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вопрос

To solve the equation 3cos^2(x) = 7(sin(x) + 1), we can use the Pythagorean identity sin^2(x) + cos^2(x) = 1 to substitute for cos^2(x) in terms of sin(x).

3(1 - sin^2(x)) = 7(sin(x) + 1)
3 - 3sin^2(x) = 7sin(x) + 7

Rearranging the terms, we get:
3sin^2(x) + 7sin(x) - 4 = 0

Now, let's solve this quadratic equation for sin(x). We can factor it, or use the quadratic formula:

sin(x) = [-7 ± √(7^2 - 4(3)(-4))] / 2(3)
sin(x) = [-7 ± √(49 + 48)] / 6
sin(x) = [-7 ± √97] / 6

Therefore, the solutions for sin(x) are:
sin(x) = (-7 + √97) / 6
sin(x) = (-7 - √97) / 6

Now that we have the values of sin(x), we can find the corresponding values of cos(x) using the Pythagorean identity:

cos(x) = ±√(1 - sin^2(x))

For the positive value of sin(x):
cos(x) = ±√(1 - (-7 + √97)^2 / 36
For the negative value of sin(x):
cos(x) = ±√(1 - (-7 - √97)^2 / 36

These are the solutions to the equation 3cos^2(x) = 7(sin(x) + 1).