Question

Sinx+cosx=2.5+5sinx*cosx

Answer 1

$sinx+cosx=2,5+5sinx\cdot cosx\\\\Zamena:\; \; t=sinx+cosx\\\\t^2=(sinx+cosx)^2=\underbrace {sin^2x+cos^2x}_{1}+2sinx\cdot cosx=1+2sinx\cdot cosx\; \to \\\\2sinx\cdot cosx=t^2-1\; ,\quad sinx\cdot cosx=\frac{t^2-1}{2}\\\\t=2,5+5\cdot \frac{t^2-1}{2}\; \; ,\; \; \; t=\frac{5}{2}+\frac{5(t^2-1)}{2}\; \; ,\; \; 2t=5+5t^2-5\; ,\\\\5t^2-2t=0\; \; ,\; \; \; t\, (5t-2)=0\; \; \to \; \; t=0\; \; ili\; \; 5t-2=0\\\\a)\; \; t=0\; \; ,\; \; \; sinx+cosx=0\, |:cosx\ne 0\; ,\\\\tgx+1=0\; \; ,\; \; tgx=-1\; ,\\\\x=-\frac{\pi}{4}+\pi n\; ,\; n\in Z$

$b)\; \; 5t-2=0\; \; ,\; \; t=\frac{2}{5}\; \; ,\; \; sinx+cosx=\frac{2}{5} \; ,\\\\sinx+sin(\frac{\pi}{2}-x)=\frac{2}{5}\\\\2sin\frac{x+(\frac{\pi}{2}+x)}{2}\cdot cos\frac{x-(\frac{\pi}{2}-x)}{2}=\frac{2}{5}\\\\2sin(x+\frac{\pi}{4})\cdot cos(-\frac{\pi}{4})=\frac{2}{5}\\\\\sqrt2\cdot sin(x+\frac{\pi}{4})=\frac{2}{5}\\\\sin(x+\frac{\pi }{4})=\frac{\sqrt2}{5}\\\\x+\frac{\pi}{4}=(-1)^{n}\cdot arcsin\frac{\sqrt2}{5}+\pi n\; ,\; n\in Z\\\\x=-\frac{\pi }{4}+(-1)^{n}\cdot arcsin\frac{\sqrt2}{5}+\pi n\; ,\; n\in Z$

$Otvet:\; \; x=-\frac{\pi}{4}+\pi n\; ,\; \; x=-\frac{\pi }{4}+(-1)^{n}\cdot arcsin\frac{\sqrt2}{5}+\pi n\; ,\; n\in Z\; .$

Answer 2

To solve the trigonometric equation $\sin(x) + \cos(x) = 2.5 + 5\sin(x)\cos(x)$, we can use some trigonometric identities to simplify it. First, let's rewrite the equation in a more manageable form.

Recall the double angle identity for sine: $\sin(2x) = 2\sin(x)\cos(x)$.

Now, let's try to rewrite the given equation using this identity:

$\sin(x) + \cos(x) = 2.5 + 5\sin(x)\cos(x)$ $\sin(x) + \cos(x) - 5\sin(x)\cos(x) = 2.5$

Next, let's try to combine the trigonometric terms on the left side:

$\sin(x) + \cos(x) - 5\sin(x)\cos(x)$

Notice that the left side looks like $\sin(x) + \cos(x)$ which suggests that we might be able to rewrite it in terms of the double angle identity. Let's try to use the double angle identity for sine:

$\sin(2x) = 2\sin(x)\cos(x)$

Now, if we divide both sides of this identity by 2:

$\frac{\sin(2x)}{2} = \sin(x)\cos(x)$

Substitute this into the equation:

$\sin(x) + \cos(x) - 5\left(\frac{\sin(2x)}{2}\right) = 2.5$

Now, we have a single trigonometric function $\sin(2x)$ instead of products of $\sin(x)$ and $\cos(x)$. Let's proceed with solving the equation:

$\sin(x) + \cos(x) - \frac{5\sin(2x)}{2} = 2.5$

Next, let's use the double angle identity for sine again:

$\sin(2x) = 2\sin(x)\cos(x)$

Now, rewrite the equation with this identity:

$\sin(x) + \cos(x) - \frac{5(2\sin(x)\cos(x))}{2} = 2.5$

Simplify the equation further:

$\sin(x) + \cos(x) - 5\sin(x)\cos(x) = 2.5$

This is exactly the same equation we started with. It appears that the equation is not easily solvable using these trigonometric identities. It's possible that there might be some other methods or approximations to find an approximate solution, but there is no simple closed-form solution for this equation.