4cos^2x+sinxcosx+3sin^2-3=0

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вопрос

To solve the equation 4cos^2x + sinxcosx + 3sin^2x - 3 = 0, we can first rewrite it using trigonometric identities.

Since cos^2x + sin^2x = 1, we can rewrite the equation as:

4(1 - sin^2x) + sinx*cosx + 3sin^2x - 3 = 0.

Now, we can simplify it further:

4 - 4sin^2x + sinxcosx + 3sin^2x - 3 = 0
4 - sin^2x + sinxcosx = 0.

Next, we can use the identity sin 2x = 2sinxcosx to rewrite the equation:

4 - sin^2x + sinx*cosx = 0
4 - sin^2x + sin 2x / 2 = 0
4 - sin^2x + sin 2x / 2 = 0
4 - (2sinx)^2 + sin 2x / 2 = 0
4 - 2sin^2x + 2sinxcosx = 0
2(2 - sin^2x + sinxcosx) = 0.

Now, we can see that the equation is simplified to:

2(2 - sinx)(1 + sinx) = 0.

This equation can be solved by setting each factor equal to zero:

2 - sinx = 0 or 1 + sinx = 0
sinx = 2 or sinx = -1.

However, the value of sin x cannot be 2, so we discard that and keep the value sinx = -1.

Therefore, the solution to the equation 4cos^2x + sinxcosx + 3sin^2x - 3 = 0 is sin x = -1.