Question

3cos^2x+2sinxcosx+5sin^2x=3

Answer 1

$3\cos^2x+2\sin x\cos x+5\sin^2x=3(\sin^2x+\cos^2x)\\\\3\cos^2x+2\sin x\cos x+5\sin^2x=3\sin^2x+3\cos^2x\\\\2\sin^2x+2\sin x\cos x=0|:2\cos^2x\\\\tg^2x+tgx=0\Rightarrow\left \ [ {{tg x=0,} \atop {tg x=-1}} \right. \\\\\left \ [ {{z=arctg(0)+\pi k,} \atop {x=arctg(-1)+\pi k}} \right. \left \ [ {{x=\pi k,} \atop {x=-\frac{\pi}{4}+\pi k }} \right.$

ОТВЕТ: $x=\pi k, k\in\mathbb Z; x=-\frac{\pi}{4}+ \pi k, k\in\mathbb Z$.

Answer 2

To solve the trigonometric equation $3\cos^2x + 2\sin x \cos x + 5\sin^2x = 3$, we can make use of the trigonometric identities to simplify the equation. The Pythagorean identity $\sin^2x + \cos^2x = 1$ will be particularly helpful here.

Step 1: Use the Pythagorean identity to simplify the equation: $3\cos^2x + 2\sin x \cos x + 5\sin^2x = 3$ $3(\cos^2x + \sin^2x) + 2\sin x \cos x = 3$ $3 + 2\sin x \cos x = 3$

Step 2: Move all terms to one side of the equation: $2\sin x \cos x = 3 - 3$ $2\sin x \cos x = 0$

Step 3: Factor out the common term: $2\sin x \cos x = 0$ $2\sin x \cdot 1 \cdot \cos x = 0$

Step 4: Apply the zero-product property: $2\sin x = 0 \quad \text{or} \quad \cos x = 0$

Step 5: Solve for $x$ in each equation:

For $2\sin x = 0$: $\sin x = 0$ The solutions for $\sin x = 0$ are $x = 0 + k\pi$ where $k$ is an integer.

For $\cos x = 0$: The solutions for $\cos x = 0$ are $x = \frac{\pi}{2} + k\pi$ where $k$ is an integer.

So, the general solutions for $x$ are: $x = 0 + k\pi \quad \text{and} \quad x = \frac{\pi}{2} + k\pi$ where $k$ is an integer.