Sinx(4 sinx−1)=2+3√cosx

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вопрос

To solve this equation, we first need to expand the left side of the equation:

Sinx(4 sinx - 1) = 4sin^2x - sinx

Next, we substitute this expansion back into the original equation:

4sin^2x - sinx = 2 + 3√cosx

Now we need to convert sinx and cosx into the same trigonometric function so that we can solve the equation. We know that sin^2x + cos^2x = 1, so cosx = 1 - sin^2x.

Substitute cosx = 1 - sin^2x into the equation:

4sin^2x - sinx = 2 + 3√(1 - sin^2x)

Now, isolate the square root term on one side of the equation:

4sin^2x - sinx - 2 = 3√(1 - sin^2x)

Square both sides of the equation to eliminate the square root:

(4sin^2x - sinx - 2)^2 = 9(1 - sin^2x)

Expand both sides:

16sin^4x - 8sin^3x - 16sin^2x + 8sinx + 4 - sin^2x = 9 - 9sin^2x

Combine like terms:

16sin^4x - 8sin^3x - 17sin^2x + 8sinx - 5 = 9 - 9sin^2x

Rearrange the terms:

16sin^4x - 8sin^3x - 8sin^2x - 8sinx - 14 = 0

This is a quartic equation in terms of sinx. Solving this equation for sinx involves finding the roots using methods such as factoring, the quadratic formula, or numerical methods.