Question

Решите тригонометрические уравнения 1)4sinx+3cosx=6 2)2sinx+cosx=2 3)sin^2x+sin^2x=sin^23x

Answer 1

$1)\; \; 4sinx+3cosx=6\; |:5\\\\ \frac{4}{5}\cdot sinx+ \frac{3}{5}\cdot cosx= \frac{6}{5}\\\\T.k.\; \; ( \frac{4}{5})^2 + (\frac{3}{5})^2= \frac{16}{25} + \frac{9}{25}=1 \; ,\; to \; \; \frac{4}{5}=cosa\; ,\; \frac{3}{5}=sina\; \to \\\\tga=\frac{3}{4} \; \; \Rightarrow \; \; a=arctg\frac{3}{4}\\\\cosa\cdot sinx+sina\cdot cosx= \frac{6}{5} \\\\sin(x+a)= \frac{6}{5} \ \textgreater \ 1\; \; \to \; \; net\; \; reshenij$

$2)\; \; 2sinx+cosx=2\; |:\sqrt5\\\\ \frac{2}{\sqrt5}\cdot sinx+ \frac{1}{\sqrt5}\cdot cosx =\frac{2}{\sqrt5} \\\\ cosa=\frac{2}{\sqrt5} \; ,\; \; sina=\frac{1}{\sqrt5} \; ,\; \; a=arctg \frac{1}{2} \\\\sin(x+a)= \frac{2}{\sqrt5} \\\\x+a=(-1)^{n}arcsin\frac{2}{\sqrt5}+\pi n,\; n\in Z\\\\x=(-1)^{n}arcsin\frac{2}{\sqrt5}-a+\pi n,\; n\in Z\\\\x=(-1)^{n}arcsin\frac{2}{\sqrt5}-arctg\frac{1}{2}+\pi n,\; n\in Z$