Question

12 cos^2x +cos2x=5sin2x+1

Answer 1

12*cos^2 x  +cos^2 x-sin^2 x=10*sinx*cosx+cos^2 x+sin^2 x
(12*cos^2 x-10sinx*cosx-2sin^2 x)=0 всё делим на 2sin^2 x
6ctgx-5ctgx-1=0
ctgx=1
x=pi/4+pi*n

Answer 2

To solve the trigonometric equation $12\cos^2(x) + \cos(2x) = 5\sin(2x) + 1$, we can use various trigonometric identities and properties to simplify and solve for $x$. Let's break down the steps:

Step 1: Use the double-angle formula for cosine: $\cos(2x) = 2\cos^2(x) - 1$. Step 2: Use the double-angle formula for sine: $\sin(2x) = 2\sin(x)\cos(x)$.

Substitute these into the original equation:

$12\cos^2(x) + (2\cos^2(x) - 1) = 5(2\sin(x)\cos(x)) + 1.$

Simplify the equation further:

$12\cos^2(x) + 2\cos^2(x) - 1 = 10\sin(x)\cos(x) + 1.$

Combine like terms:

$14\cos^2(x) - 10\sin(x)\cos(x) = 2.$

Divide the entire equation by $2$ to simplify:

$7\cos^2(x) - 5\sin(x)\cos(x) = 1.$

Now, we can use the identity $\sin(2x) = 2\sin(x)\cos(x)$ to substitute for $5\sin(x)\cos(x)$:

$7\cos^2(x) - \frac{1}{2}\sin(2x) = 1.$

At this point, we have an equation in terms of both $\cos(x)$ and $\sin(2x)$. Unfortunately, it's not immediately clear how to solve for $x$ directly.

However, if you have a specific range of values for $x$ (e.g., $0 \leq x \leq 2\pi$), you can use a numerical approach or graphing software to find approximate solutions.

If you're looking for a general solution in terms of $x$, it might not be possible due to the complex interaction between sine and cosine functions in the equation. It could involve transcendental equations which don't have a simple algebraic solution.

If you have more context or specific requirements, please provide additional information so I can assist you further.