Question

1-8sinx*cosx+6cosx^2=0

Answer 1

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Answer 2

To solve the equation 1 - 8sin(x)cos(x) + 6cos^2(x) = 0, we can use some trigonometric identities and algebraic manipulations. Let's break it down step by step:

1. Start by observing that 1 - 8sin(x)cos(x) + 6cos^2(x) = 0 is a quadratic equation in terms of cos(x). We can rearrange it as follows:

   6cos^2(x) - 8sin(x)cos(x) + 1 = 0

2. We notice that this equation can be simplified using the Pythagorean identity sin^2(x) + cos^2(x) = 1. Rearrange it to isolate sin(x) and substitute it into the quadratic equation:

   sin^2(x) = 1 - cos^2(x) sin(x) = √(1 - cos^2(x))

   6cos^2(x) - 8sin(x)cos(x) + 1 = 0 6cos^2(x) - 8√(1 - cos^2(x))cos(x) + 1 = 0

3. Let's substitute u = cos(x) to simplify the equation further:

   6u^2 - 8√(1 - u^2)u + 1 = 0

4. Now, we have a quadratic equation in terms of u. We can solve it using the quadratic formula:

   u = [-(-8√(1 - u^2)) ± √((-8√(1 - u^2))^2 - 4(6)(1))]/(2(6)) u = [8√(1 - u^2) ± √(64(1 - u^2) - 24)]/(12) u = [8√(1 - u^2) ± √(64 - 64u^2 - 24)]/(12) u = [8√(1 - u^2) ± √(40 - 64u^2)]/(12)

5. Simplify the expression further:

   u = [2√(1 - u^2) ± √(10 - 16u^2)]/(3)

6. At this point, we have a transcendental equation, and it's challenging to find exact solutions algebraically. We can use numerical methods or approximation techniques to estimate the values of u and subsequently cos(x). One such method is the Newton-Raphson method, which can be used to iteratively refine the solutions.

   However, if you're looking for specific values or a more accurate solution, please provide any constraints or specific range for x so that we can assist you further.