2sin^2x-5sinxcosx+3cos^2x=0

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вопрос

To solve the equation 2sin^2x - 5sinxcosx + 3cos^2x = 0, we can use the trigonometric identity sin^2x + cos^2x = 1 to simplify the equation.

Starting with the given equation:
2sin^2x - 5sinxcosx + 3cos^2x = 0

We know that sin^2x + cos^2x = 1, so we can substitute cos^2x = 1 - sin^2x into the equation:
2sin^2x - 5sinx(1 - sin^2x) + 3(1 - sin^2x) = 0
2sin^2x - 5sinx + 5sin^3x + 3 - 3sin^2x = 0
Rearranging the terms:
5sin^3x - 3sin^2x + 2sin^2x - 5sinx + 3 = 0
5sin^3x - sin^2x - 5sinx + 3 = 0

Now we have a cubic equation in terms of sinx. This equation can be solved using various methods such as factoring, substitution, or numerical methods. Once the values for sinx are found, the corresponding values for x can be determined.