2 sin^2x+ 3√2 cos(3π/2+x)+2=0

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To solve the equation (2 \sin^2x + 3\sqrt{2} \cos\left(\frac{3\pi}{2}+x)+2=0), we can simplify it using trigonometric identities.

First, note that (\cos\left(\frac{3\pi}{2}+x)) can be rewritten using the angle addition formula for cosine:

(\cos\left(\frac{3\pi}{2}+x)= \cos\frac{3\pi}{2} \cos x - \sin\frac{3\pi}{2}\sin x = 0*(-\sin x) - (-1) \cdot \cos x = \cos x).

Therefore, the equation simplifies to:

(2\sin^2 x + 3\sqrt{2}\cos x + 2 = 0)

Now, recall that (\sin^2 x + \cos^2 x = 1). We can rewrite the equation in terms of either sine or cosine:

(2(1 - \cos^2 x) + 3\sqrt{2}\cos x + 2 = 0)

(2 - 2\cos^2 x + 3\sqrt{2}\cos x + 2 = 0)

(-2\cos^2 x + 3\sqrt{2}\cos x + 4 = 0)

We can now treat this as a quadratic equation in terms of cosine. Let (u = \cos x) to simplify the equation:

(-2u^2 + 3\sqrt{2}u + 4 = 0)

Now, we can solve this quadratic equation for u using the quadratic formula:

(u = \frac{-b ± \sqrt{b^2 - 4ac}}{2a})

Substitute (a = -2), (b = 3\sqrt{2}), and (c = 4) into the formula:

(u = \frac{-3\sqrt{2} ± \sqrt{(3\sqrt{2})^2 - 4(-2)4}}{2*(-2)})

(u = \frac{-3\sqrt{2} ± \sqrt{18 - 32}}{-4})

(u = \frac{-3\sqrt{2} ± \sqrt{-14}}{-4})

Since the square root of -14 is imaginary, the solutions for u are complex. This means that there are no real solutions for the original trigonometric equation.