Question

Sin^2x+2sin2x+3cos^2x=0

Answer 1

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Answer 2

Solving the Equation Sin^2x + 2sin2x + 3cos^2x = 0

To solve the equation Sin^2x + 2sin2x + 3cos^2x = 0, we can simplify it by using trigonometric identities and then solve for x.

Let's break down the steps:

1. Use the identity sin^2x + cos^2x = 1 to rewrite the equation: - Sin^2x + 2sin2x + 3cos^2x = 0 - Sin^2x + 2sin2x + 3(1 - sin^2x) = 0 - Sin^2x + 2sin2x + 3 - 3sin^2x = 0 - -2sin^2x + 2sin2x + 3 = 0

2. Rearrange the terms: - 2sin2x - 2sin^2x + 3 = 0

Now, let's solve the equation:

To solve the equation 2sin2x - 2sin^2x + 3 = 0, we can factor out a common factor of 2:

2(sin2x - sin^2x) + 3 = 0

Next, we can use the identity sin2x = 2sinxcosx to simplify further:

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

Now, we have a quadratic equation in terms of sinx. Let's solve it:

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

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

To solve this quadratic equation, we can use the quadratic formula:

sinx = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = -2, b = 4cosx, and c = 3.

Plugging in the values, we get:

sinx = (-4cosx ± √((4cosx)^2 - 4(-2)(3))) / (2(-2))

Simplifying further:

sinx = (-4cosx ± √(16cos^2x + 24)) / (-4)

sinx = (cosx ± √(cos^2x + 6)) / 2

Therefore, the solutions for x are:

x = arcsin((cosx + √(cos^2x + 6)) / 2) + kπ

x = arcsin((cosx - √(cos^2x + 6)) / 2) + kπ

where k is an integer.

Please note that the above solution is derived using trigonometric identities and the quadratic formula. It is always a good practice to verify the solution by substituting it back into the original equation to ensure its validity.

I hope this helps! Let me know if you have any further questions.