Question

Sin^2x+3sinxcosx-4cos^2x=0

Answer 1

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Answer 2

To solve the equation sin^2(x) + 3sin(x)cos(x) - 4cos^2(x) = 0, we can use some trigonometric identities and algebraic manipulation.

Let's break it down step by step:

1. Start by using the identity sin^2(x) + cos^2(x) = 1 to replace sin^2(x) with 1 - cos^2(x).

The equation becomes: (1 - cos^2(x)) + 3sin(x)cos(x) - 4cos^2(x) = 0

2. Distribute the 3sin(x)cos(x) term: 1 - cos^2(x) + 3sin(x)cos(x) - 4cos^2(x) = 0

3. Rearrange the terms: -5cos^2(x) + 3sin(x)cos(x) + 1 = 0

4. Factor out a common factor of cos(x): cos(x)(-5cos(x) + 3sin(x)) + 1 = 0

5. Now we have two cases to consider:

Case 1: cos(x) = 0

If cos(x) = 0, then x is an angle where the cosine function equals zero. This occurs at x = π/2 + nπ and x = 3π/2 + nπ, where n is an integer.

Case 2: -5cos(x) + 3sin(x) = -1

To solve this equation, we can use the Pythagorean identity sin^2(x) + cos^2(x) = 1. Rearranging, we get sin^2(x) = 1 - cos^2(x).

Substituting this into the equation, we have: -5cos(x) + 3(√(1 - sin^2(x))) = -1

Simplifying further: -5cos(x) + 3√(1 - sin^2(x)) = -1

Squaring both sides: 25cos^2(x) - 6cos(x)√(1 - sin^2(x)) + 9(1 - sin^2(x)) = 1

Using the Pythagorean identity sin^2(x) + cos^2(x) = 1, we can simplify the equation to: 25cos^2(x) - 6cos(x)√(cos^2(x)) + 9(1 - cos^2(x)) = 1

25cos^2(x) - 6cos^2(x) + 9 - 9cos^2(x) = 1

10cos^2(x) - 9 = 0

10cos^2(x) = 9

cos^2(x) = 9/10

Taking the square root of both sides: cos(x) = ±√(9/10)

Now, we have two cases to consider for the value of cos(x):

Case 2.1: cos(x) = √(9/10)

To solve for x, we take the inverse cosine of both sides: x = arccos(√(9/10))

Case 2.2: cos(x) = -√(9/10)

To solve for x, we take the inverse cosine of both sides: x = arccos(-√(9/10))

Please note that both cases should be checked in the original equation to confirm whether they satisfy the given equation.