(2sin^2x-sin2x-2cos2x)/sqrt(1-x^2)=0

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вопрос

To solve this equation, we will first simplify the expression in the numerator:

Given expression: (2sin^2x - sin2x - 2cos2x)

Using the double-angle identity for sine and cosine:

sin2x = 2sinx*cosx
cos2x = 2cos^2x - 1

Therefore, the expression becomes:

2sin^2x - sin2x - 2cos2x
= 2sin^2x - 2sinxcosx - 2(2cos^2x - 1)
= 2sin^2x - 2sinxcosx - 4cos^2x + 2

Now, the expression becomes: 2sin^2x - 2sinx*cosx - 4cos^2x + 2

Therefore, the original equation becomes:

(2sin^2x - 2sinx*cosx - 4cos^2x + 2) / sqrt(1 - x^2) = 0

Now we need to solve the equation:

2sin^2x - 2sinx*cosx - 4cos^2x + 2 = 0

Dividing by 2:

sin^2x - sinx*cosx - 2cos^2x + 1 = 0

Converting sinxcosx to (1/2)sin2x:

sin^2x - (1/2)*sin2x - 2cos^2x + 1 = 0

Rearranging the equation:

sin^2x - (1/2)*sin2x - 4cos^2x + 1 = 0

This equation can be further simplified using trigonometric identities to solve for x.