Question

7cos2x-4sin2x=-4 Как решить ?

Answer 1

7cos2x-4sin2x= - 4

Решаем как однородное.

7(Сos²x - Sin²x) - 4*2SinxCosx  = -4*1

7Cos²x - 7Sin²x -8SinxCosx = -4(Sin²x + Cos²x)

7Cos²x - 7Sin²x -8SinxCosx  +4Sin²x + 4Cos²x=0

11Cos²x -8SinxCosx -3Sin²x = 0 | :Сos²x≠ 0

11 -8tg²x -3tg²x = 0

tgx = t

-3t² -8t +11 = 0

3t² +8t -11 = 0

t = (-4+-√(16 +33))/3 = (-4 +-7)/3

t₁ = -11/3                                    t₂ = 1

tgx = -11/3                                 tgx = 1

x = -arctg11/3 + πk , k ∈Z         x = π/4 + πn , n ∈ Z

Answer 2

$7cos2x-4sin2x=-4\; |:\sqrt{65}\\\\\frac{7}{\sqrt{65}}\cdot cos2x-\frac{4}{\sqrt{65}}\cdot sin2x=-\frac{4}{\sqrt{65}}\\\\(\frac{7}{\sqrt{65}})^2+(\frac{4}{\sqrt{65}})^2=1\; \; \to \; \; \frac{7}{\sqrt{65}}=sina\; ,\; \; \frac{4}{\sqrt{65}}=cosa\; \; \to \; \; tga=\frac{7}{4}\; ,\\\\a=arctg\frac{7}{4}\\\\sina\cdot cos2x-cosa\cdot sin2x=-\frac{4}{\sqrt{65}}\\\\sin(a-2x)=-\frac{4}{\sqrt{65}}\\\\a-2x=(-1)^{n}\cdot arcsin(-\frac{4}{\sqrt{65}})+\pi n=(-1)^{n+1}\cdot arcsin\frac{4}{\sqrt{65}}+\pi n\; ,\; n\in Z\\\\2x=a-(-1)^{n+1}\cdot arcsin\frac{4}{\sqrt{65}}-\pi n\; ,\; n\in Z\\\\x=\frac{1}{2}\cdot (arctg\frac{7}{4}+(-1)^{n+2}\cdot arcsin\frac{4}{\sqrt{65}}+\pi n)\; ,\; n\in Z$