Sin^2x -2sinxcosx -3cos^2x=0

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To solve this trigonometric equation, we can use the Pythagorean identity sin^2x + cos^2x = 1.

We can rewrite the equation as:

sin^2x - 2sinxcosx - 3(1 - sin^2x) = 0

Expanding and simplifying, we get:

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

Let's rewrite the equation as:

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

Let's factor the equation as follows:

(2sinx + 1)(2sinx - 3) = 0

Setting each factor to zero gives:

2sinx + 1 = 0
sinx = -1/2
x = arcsin(-1/2) + 2nπ or x = π - π/6 + 2nπ or x = 5π/6 + 2nπ

2sinx - 3 = 0
sinx = 3/2 (which is not a valid solution since sine function has a range of -1 to 1)

So, the solutions to the equation are x = π - π/6 + 2nπ or x = 5π/6 + 2nπ, where n is an integer.