2 cos²x + 5 sinx - 4 = 0

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To solve this quadratic equation, we can first notice that there are two different trigonometric functions present: cosine and sine. We can use the trigonometric identity that relates cosine squared and sine squared.

The identity that relates cosine squared and sine squared is:
cos^2(x) + sin^2(x) = 1

Let's re-write the given equation in terms of either cosine squared or sine squared:

2(1 - sin^2(x)) + 5sin(x) - 4 = 0
2 - 2sin^2(x) + 5sin(x) - 4 = 0
-2sin^2(x) + 5sin(x) - 2 = 0

This is a quadratic equation in terms of sin(x). Let's substitute sin(x) = y:

-2y^2 + 5y - 2 = 0

Now, we can solve this quadratic equation for y using factoring, quadratic formula or completing the square. Let's try factoring:

(-2y + 1)(y - 2) = 0

Setting each factor to zero and solving for y gives:

-2y + 1 = 0 or y - 2 = 0
y = 1/2 or y = 2

Since sin(x) = y:

sin(x) = 1/2 or sin(x) = 2

Since sin(x) cannot be greater than 1 or less than -1, the solution is:

sin(x) = 1/2

To determine all solutions within the interval [0, 2π], we know that sin(π/6) = 1/2. So, x = π/6 or 30 degrees.

Therefore, the solution to the equation 2cos^2x + 5sinx - 4 = 0 is x = π/6 or x = 30 degrees.