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 equation $\sin(x) + \cos(x) = 2.5 + 5\sin(x)\cos(x)$, we can start by manipulating the equation to get it into a more workable form. Let's proceed step by step:

Step 1: Rearrange the equation to gather all the terms on one side: $\sin(x) + \cos(x) - 2.5 - 5\sin(x)\cos(x) = 0$

Step 2: Try to simplify the left-hand side of the equation. We can use the trigonometric identity $\sin^2(x) + \cos^2(x) = 1$: $\sin(x) + \cos(x) - 2.5 - 5\sin(x)\cos(x) = (\sin^2(x) + \cos^2(x)) - 5\sin(x)\cos(x) - 2.5$

Step 3: Now, we can use the double angle formula for cosine to express $2\cos(x) \sin(x)$ in terms of a single trigonometric function: $2\cos(x)\sin(x) = \sin(2x)$

Step 4: Apply the above result to the equation: $(\sin^2(x) + \cos^2(x)) - 5\sin(x)\cos(x) - 2.5 = 1 - 5\sin(2x) - 2.5$

Step 5: Simplify further: $1 - 5\sin(2x) - 2.5 = -1 - 5\sin(2x)$

Step 6: Now, we have the equation: $-1 - 5\sin(2x) = 0$

Step 7: To find the solutions for $x$, solve for $\sin(2x)$: $\sin(2x) = -\frac{1}{5}$

Step 8: We know that $-1 \leq \sin(2x) \leq 1$ for all values of $x$. Since $-\frac{1}{5}$ is outside this range, there are no real solutions for $x$ that satisfy the original equation.

Therefore, the equation $\sin(x) + \cos(x) = 2.5 + 5\sin(x)\cos(x)$ has no real solutions.