Question

Giải các phương trình sau :

a) \(3sin^2x-4sinxcosx+5cos^2x=2\)

b) \(25sin^2x+15sin2x+9cos^2=25\)

c) sinx + cosx =1

d) 3cos2x - 4sin2x =1

f) \(4sin^2x-6cos^2x=0\)

g) \(5sin2x-6cos^2x=13\)

h) \(sinx=\sqrt{3}cosx\)

i) \(sin^4x+cos^4\left(x+\frac{\pi}{4}\right)=\frac{1}{4}\)

j)\(tanx+2cotx-3=0\)

k) \(tan^25x=\frac{1}{3}\)

m) \(sin^4x-cos^4x=cosx-2\)

Answer 1

a)

PT $\Leftrightarrow \sin ^2x-4\sin x\cos x+3\cos ^2x+2(\sin ^2x+\cos ^2x)=2$

$\Leftrightarrow \sin ^2x-4\sin x\cos x+3\cos ^2x=0$

$\Leftrightarrow (\sin x-3\cos x)(\sin x-\cos x)=0$

Nếu $\sin x-3\cos x=0$. Dễ thấy $\sin x, \cos x\neq 0$ nên $\tan x=\frac{\sin x}{\cos x}=3$

$\Rightarrow x=k\pi +\tan ^{-1}(3)$ với $k$ nguyên

Nếu $\sin x=\cos x$ thì tương tự ta có $\tan x=1\Rightarrow x=\pi (k+\frac{1}{4})$ với $k$ nguyên

Answer 2

b)
PT $\Leftrightarrow 25(\sin ^2x+\cos ^2x)+30\sin x\cos x-16\cos ^2x=25$

$\Leftrightarrow 30\sin x\cos x-16\cos ^2x=0$

$\Leftrightarrow \cos x(15\sin x-8\cos x)=0$

Nếu $\cos x=0\Rightarrow x=\pi (k+\frac{1}{2})$ với $k$ nguyên

Nếu $15\sin x-8\cos x=0$

Dễ thấy $\cos x\neq 0$ nên suy ra $\tan x=\frac{\sin x}{\cos x}=\frac{8}{15}$

$\Rightarrow x=k\pi +\tan ^{-1}(\frac{8}{15})$ với $k$ nguyên

Answer 3

c) \(\left\{\begin{matrix} \sin x+\cos x=1\\ \sin ^2x+\cos ^2x=1\end{matrix}\right.\Rightarrow \left\{\begin{matrix} (\sin x+\cos x)^2=1\\ \sin ^2x+\cos ^2x=1\end{matrix}\right.\)

\(\Rightarrow 2\sin x\cos x=0\Leftrightarrow \sin 2x=0\Rightarrow x=\frac{k}{2}\pi\) với $k$ nguyên.

Answer 4

d) \(\left\{\begin{matrix} 3\cos 2x-4\sin 2x=1\\ \sin ^22x+\cos ^22x=1\end{matrix}\right.\) \(\Rightarrow \left\{\begin{matrix} \sin 2x=\frac{3\cos 2x-1}{4}\\ \sin ^22x+\cos ^22x=1\end{matrix}\right.\)

\(\Rightarrow (\frac{3\cos 2x-1}{4})^2+\cos ^22x=1\)

\(\Leftrightarrow 25\cos ^22x-6\cos 2x-15=0\Rightarrow \cos 2x=\frac{3\pm 8\sqrt{6}}{25}\)

Mà $\cos 2x=\frac{1+4\sin 2x}{3}\in [\frac{1}{3}; \frac{5}{3}]$ nên $\cos 2x=\frac{3+8\sqrt{6}}{25}$

$\Rightarrow x=\pm \frac{1}{2}\cos ^{-1}\frac{3+8\sqrt{6}}{25}+k\pi$ với $k$ nguyên.

Answer 5

f)

$4\sin ^2x-6\cos ^2x=0$

$\Leftrightarrow (2\sin x-\sqrt{6}\cos x)(2\sin x+\sqrt{6}\cos x)=0$

Nếu $2\sin x-\sqrt{6}\cos x=0$

Dễ thấy $\cos x\neq 0$ nên $\tan x=\frac{\sin x}{\cos x}=\frac{\sqrt{6}}{2}$

$\Rightarrow x=k\pi +\tan ^{-1}\frac{\sqrt{6}}{2}$ với $k$ nguyên

Tương tự $2\sin x+\sqrt{6}\cos x=0\Rightarrow x=k\pi +\tan ^{-1}\frac{-\sqrt{6}}{2}$ với $k$ nguyên

Answer 6

g)

$5\sin 2x-6\cos ^2x=13$

Ta thấy: $5\sin 2x\leq 5$ với mọi $x\in\mathbb{R}$

$\cos ^2x\geq 0$ với mọi $x\in\mathbb{R}$

Do đó: $5\sin 2x-6\cos ^2x\leq 5 < 13$ với mọi $x\in\mathbb{R}$

Do đó pt vô nghiệm.

Answer 7

h)

$\sin x=\sqrt{3}\cos x$

Nếu $\cos x=0\Rightarrow \sin x=0\Rightarrow \sin ^2x+\cos ^2x=0$ (vô lý)

Do đó $\cos x\neq 0$

$\Rightarrow \tan x=\frac{\sin x}{\cos x}=\sqrt{3}$

$\Rightarrow x=k\pi +\tan ^{-1}\sqrt{3}$ với $k$ nguyên.

Answer 8

i)

$\sin ^4x+\cos ^4(x+\frac{\pi}{4})=\frac{1}{4}$

$\Leftrightarrow \sin ^4x+(\frac{\sqrt{2}}{2}\cos x-\frac{\sqrt{2}}{2}\sin x)^4=\frac{1}{4}$

$\Leftrightarrow 4\sin ^4x+(\cos x-\sin x)^4=1$

$\Leftrightarrow 4\sin ^4x+(1-2\sin x\cos x)^2=1$

$\Leftrightarrow 4\sin ^2x(1-\cos ^2x)+4\sin ^2x\cos ^2x-4\sin x\cos x=0$

$\Leftrightarrow 4\sin ^2x-4\sin x\cos x=0$

$\Leftrightarrow \sin x(\sin x-\cos x)=0$

Nếu $\sin x=0\Rightarrow x=k\pi$ với $k$ nguyên

Nếu $\sin x=\cos x\Rightarrow \tan x=1\Rightarrow x=\pi (k+\frac{1}{4})$ với $k$ nguyên.

Answer 9

j) ĐK:$x\neq \frac{k\pi}{2}$ với $k$ nguyên.

$\tan x+2\cot x-3=0$

$\Leftrightarrow \tan x+\frac{2}{\tan x}-3=0$

$\Leftrightarrow \tan ^2x-3\tan x+2=0$

$\Leftrightarrow (\tan x-1)(\tan x-2)=0$

Nếu $\tan x=1\Rightarrow x=\pi (k+\frac{1}{4})$ với $k$ nguyên.

Nếu $\tan x=2\Rightarrow x=k\pi +\tan ^{-1}2$ với $k$ nguyên.

Answer 10

k) ĐK:.......

$\tan ^25x=\frac{1}{3}\Rightarrow \tan 5x=\pm \sqrt{\frac{1}{3}}$

$\Rightarrow 5x=k\pi +\tan ^{-1}\frac{\pm 1}{\sqrt{3}}$

$\Rightarrow x=frac{k}{5}\pi +\tan ^{-1}\frac{\pm 1}{\sqrt{3}}$ với $k$ nguyên.

Số đẹp hơn thì có thể giải như sau:

$PT \Leftrightarrow \frac{\sin ^25x}{\cos ^25x}=\frac{1}{3}$

$\Rightarrow 3\sin ^25x=\cos ^25x$

$\Rightarrow 4\\sin ^25x=1\Rightarrow \sin 5x=\pm \frac{1}{2}$

$\Rightarrow x=\frac{k\pi}{5}\pm \frac{\pi}{30}$ với $k$ nguyên.

Answer 11

m)

$\sin 4x-\cos ^4x=\cos x-2$

$\Leftrightarrow (\sin ^2x+\cos ^2x)(\sin ^2x-\cos ^2x)=\cos x-2$

$\Leftrightarrow \sin ^2x-\cos ^2x=\cos x-2$

$\Leftrightarrow 1-2\cos ^2x=\cos x-2$

$\Leftrightarrow 2\cos ^2x+\cos x-3=0$

$\Leftrightarrow (2\cos x+3)(\cos x-1)=0$

Nếu $2\cos x+3=0\Rightarrow \cos x=\frac{-3}{2}< -1$ (loại)

Nếu $\cos x-1=0\Rightarrow \cos x=1\Rightarrow x=2k\pi$ với $k$ nguyên