4^3cos^2 x-2sinx-2 =1

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Let's first simplify the equation by expanding the trigonometric term:

4^3cos^2 x - 2sinx - 2 = 1
64cos^2 x - 2sinx - 2 = 1

Next, we can use the trigonometric identity: sin^2(x) + cos^2(x) = 1 to replace cos^2(x) with 1 - sin^2(x):

64(1 - sin^2 x) - 2sinx - 2 = 1
64 - 64sin^2 x - 2sinx - 2 = 1

Rearrange the terms:

-64sin^2 x - 2sinx = -65
64sin^2 x + 2sinx = 65

Now, we have a quadratic equation in terms of sin(x). To solve for sin(x), we can use the quadratic formula:

sin(x) = (-b ± √(b^2 - 4ac)) / 2a
sin(x) = (-(2) ± √ ((2)^2 - 4(64)(-65))) / 2(64)
sin(x) = (-2 ± √(4 + 16640)) / 128
sin(x) = (-2 ± √(16644)) / 128
sin(x) = (-2 ± 129) / 128

This gives us two possible solutions for sin(x):

sin(x) = 127 / 128
sin(x) = -131 / 128

However, since the range of sin(x) is -1 to 1, the second solution is not valid. Therefore, sin(x) = 127 / 128.

Finally, we can find cos(x) using the trigonometric identity sin^2(x) + cos^2(x) = 1:

cos^2(x) = 1 - sin^2(x)
cos^2(x) = 1 - (127/128)^2
cos^2(x) = 1 - 16129 / 16384
cos^2(x) = 2555 / 16384

Taking the square root of both sides gives us:

cos(x) = ±√(2555 / 16384)

Therefore, the solutions to the equation are sin(x) = 127 / 128 and cos(x) = ±√(2555 / 16384).