Question

[Ngô Chí Thành]

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

[Akai Haruma]

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

Answer 2

[Akai Haruma]

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.