Question

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

Answer 1

Вроде вот так:
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Answer 2

To solve the trigonometric equation $6\sin^2(x) + \frac{1}{2}\sin(2x) - \cos^2(x) = 2$, we can start by using trigonometric identities to simplify the equation. Let's break it down step by step:

1. Recall the double-angle identities:

   • $\sin(2x) = 2\sin(x)\cos(x)$
   • $\cos(2x) = \cos^2(x) - \sin^2(x)$

2. Substitute the above identities into the equation: $6\sin^2(x) + \frac{1}{2}(2\sin(x)\cos(x)) - (\cos^2(x) - \sin^2(x)) = 2$

3. Distribute and simplify: $6\sin^2(x) + \sin(x)\cos(x) - \cos^2(x) + \sin^2(x) = 2$

4. Combine like terms: $7\sin^2(x) + \sin(x)\cos(x) - \cos^2(x) = 2$

5. Use the Pythagorean identity $\sin^2(x) + \cos^2(x) = 1$ to substitute for $\sin^2(x)$: $7(1 - \cos^2(x)) + \sin(x)\cos(x) - \cos^2(x) = 2$

6. Distribute and simplify again: $7 - 7\cos^2(x) + \sin(x)\cos(x) - \cos^2(x) = 2$

7. Rearrange the equation: $7 - 8\cos^2(x) + \sin(x)\cos(x) = 2$

8. Move all terms to one side of the equation: $8\cos^2(x) - \sin(x)\cos(x) + 5 = 0$

At this point, we have transformed the original equation into a quadratic equation involving $\cos(x)$. Unfortunately, solving this equation analytically for $\cos(x)$ is quite complex due to the presence of both quadratic and linear terms involving trigonometric functions.

If you have a specific numerical range for the solution or are allowed to use a numerical solver, you could use computational methods or software to find approximate solutions for $x$ that satisfy the equation.