Sin^2x-3sin×cosx+2cos^2x=0

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To solve this trigonometric equation, we will first use the Pythagorean identity:

sin^2 x + cos^2 x = 1

Next, we can rewrite the given equation using this identity:

sin^2 x - 3sin x cos x + 2cos^2 x = 0
sin^2 x + cos^2 x - 3sin x cos x = 1 - 3sin x cos x

Now, we can substitute this expression back into the equation:

1 - 3sin x cos x = 0

Adding 3sin x cos x to both sides:

1 = 3sin x cos x

Dividing both sides by 3:

sin x cos x = 1/3

Now, we can solve for x. Since sin x cos x = 1/3, we can rewrite it as:

sin x = 1/√3
cos x = √3/3

Therefore, x = π/6 + 2nπ or x = 5π/6 + 2nπ, where n is an integer.