Lời giải:

a)

\(\frac{1-\cos x}{\sin x}=\frac{(1-\cos x)(1+\cos x)}{\sin x(1+\cos x)}=\frac{1-\cos ^2x}{\sin x(1+\cos x)}=\frac{\sin ^2x}{\sin x(1+\cos x)}=\frac{\sin x}{1+\cos x}\)

b)

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

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

c)

\(\frac{\sin ^2x+2\cos x-1}{2+\cos x-\cos ^2x}=\frac{1-\cos ^2x+2\cos x-1}{2+\cos x-\cos ^2x}=\frac{-\cos ^2x+2\cos x}{2+\cos x-\cos ^2x}\)

\(=\frac{\cos x(2-\cos x)}{(2-\cos x)(\cos x+1)}=\frac{\cos x}{\cos x+1}\)

d)

\(\frac{\cos ^2x-\sin ^2x}{\cot ^2x-\tan ^2x}=\frac{\cos ^2x-\sin ^2x}{\frac{\cos ^2x}{\sin ^2x}-\frac{\sin ^2x}{\cos ^2x}}=\frac{\sin ^2x\cos ^2x(\cos ^2x-\sin ^2x)}{\cos ^4x-\sin ^4x}\)

\(=\frac{\sin ^2x\cos ^2x(\cos ^2x-\sin ^2x)}{(\cos ^2x-\sin ^2x)(\cos ^2x+\sin ^2x)}=\frac{\sin ^2x\cos ^2x}{\sin ^2x+\cos ^2x}=\sin ^2x\cos ^2x\)

e)

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

\(=\frac{\sin ^2x-\cos ^2x}{\sin ^4x}=\frac{\sin ^2x-(1-\sin ^2x)}{\sin ^4x}=\frac{2\sin ^2x-1}{\sin ^4x}=\frac{2}{\sin ^2x}-\frac{1}{\sin ^4x}\)

Ta có ddpcm.

1. 

ĐKXĐ: \(x\ne k\pi\)

\(\Leftrightarrow\left(2cos2x-1\right)\left(sinx-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=\dfrac{1}{2}\\sinx=3>1\left(ktm\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\pi}{3}+k2\pi\\2x=-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k\pi\\x=-\dfrac{\pi}{6}+k\pi\end{matrix}\right.\)

2. Bạn kiểm tra lại đề, pt này về cơ bản ko giải được.

3.

ĐKXĐ: \(x\ne\dfrac{k\pi}{2}\)

\(\dfrac{3\left(sinx+\dfrac{sinx}{cosx}\right)}{\dfrac{sinx}{cosx}-sinx}-2cosx=2\)

\(\Leftrightarrow\dfrac{3\left(1+cosx\right)}{1-cosx}+2\left(1+cosx\right)=0\)

\(\Leftrightarrow\left(1+cosx\right)\left(\dfrac{3}{1-cosx}+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=-1\left(loại\right)\\cosx=\dfrac{5}{2}\left(loại\right)\end{matrix}\right.\)

Vậy pt đã cho vô nghiệm

4.

\(\Leftrightarrow\left(sin^2x+cos^2x+2sinx.cosx\right)+\left(sinx+cosx\right)+\left(cos^2x-sin^2x\right)=0\)

\(\Leftrightarrow\left(sinx+cosx\right)^2+\left(sinx+cosx\right)+\left(sinx+cosx\right)\left(cosx-sinx\right)=0\)

\(\Leftrightarrow\left(sinx+cosx\right)\left(sinx+cosx+1+cosx-sinx\right)=0\)

\(\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)\left(2cosx+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\dfrac{\pi}{4}\right)=0\\cosx=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+k\pi\\x=\dfrac{2\pi}{3}+k2\pi\\x=-\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)

a) \(\left(sinx+cosx\right)^2=sin^2x+2sinxcosx+cos^2x\)\(=1+2sinxcosx\).
b) \(\left(sinx-cosx\right)^2=sin^2x-2sinxcosx+cos^2x\)\(=1-2sinxcosx\).
c) \(sin^4x+cos^4x=\left(sin^2x+cos^2x\right)^2-2sin^2xcos^2x\)
\(=1-2sin^2xcos^2x\).

1.

Kiểm tra lại đề bài, câu này phải là \(\dfrac{sinx+2cosx+3}{2sinx+cosx+3}\) mới đúng

2.a

ĐKXĐ: \(cosx\ne0\)

\(\Leftrightarrow\dfrac{1}{cos^2x}=4tanx+6\)

\(\Leftrightarrow1+tan^2x=4tanx+6\)

\(\Leftrightarrow tan^2x-4tanx-5=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\tanx=5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+k\pi\\x=arctan\left(5\right)+k\pi\end{matrix}\right.\)

2b.

Đặt \(x-\dfrac{\pi}{4}=t\Rightarrow x=t+\dfrac{\pi}{4}\)

\(sin^3t=\sqrt{2}sin\left(t+\dfrac{\pi}{4}\right)\)

\(\Leftrightarrow sin^3t=sint+cost\)

\(\Leftrightarrow sint\left(1-cos^2t\right)=sint+cost\)

\(\Leftrightarrow sint.cos^2t+cost=0\)

\(\Leftrightarrow cost\left(sint.cost+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cost=0\\sin2t=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}cos\left(x-\dfrac{\pi}{4}\right)=0\\sin\left(2x-\dfrac{\pi}{2}\right)=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}cos\left(x-\dfrac{\pi}{4}\right)=0\\cos2x=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow...\)

2c.

ĐKXĐ: \(sin4x\ne0\Leftrightarrow x\ne\dfrac{k\pi}{4}\)

\(\dfrac{4sinx.cos2x}{sin4x}+\dfrac{2cos2x}{sin4x}=\dfrac{2}{sin4x}\)

\(\Leftrightarrow2sinx.cos2x+cos2x=1\)

\(\Leftrightarrow2sinx.cos2x+1-2sin^2x=1\)

\(\Leftrightarrow sinx\left(cos2x-sinx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\left(loại\right)\\cos2x-sinx=0\end{matrix}\right.\)

\(\Leftrightarrow1-2sin^2x-sinx=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=-1\left(loại\right)\\sinx=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow x=\dfrac{\pi}{6}+k2\pi\)

a) Ta có: \(1-\frac{\sin^2x}{1+\cot x}-\frac{\cos^2x}{1+\tan x}=1-\frac{\sin^2x}{1+\frac{\cos x}{\sin x}}-\frac{\cos^2x}{1+\frac{\sin x}{\cos x}}\)  (Đk: sinx và cosx khác 0)

\(=1-\frac{\sin^3x}{\sin x+\cos x}-\frac{\cos^3x}{\cos x+\sin x}\)

\(=1-\frac{\left(\sin x+\cos x\right)\left(\sin^2x-\sin x.\cos x+\cos^2x\right)}{\sin x+\cos x}\)

\(=1-\left(\sin^2x+\cos^2x-\sin x.\cos x\right)\) (do sinx + cosx luôn khác 0)

\(=\sin x.\cos x\) ( do \(\sin^2x+\cos^2x=1\))

b) Ta có: \(\frac{\sin^2x+2\cos x-1}{2+\cos x-\cos^2x}=\frac{\left(\sin^2x-1\right)+2\cos x}{-\left(\cos x+1\right)\left(\cos x-2\right)}\) (Đk: cosx khác -1 và 2)

\(=\frac{-\cos x\left(\cos x-2\right)}{-\left(\cos x+1\right)\left(\cos x-2\right)}\)

\(=\frac{\cos x}{1+\cos x}\)

a) Ta có: 1−sin2x1+cotx −cos2x1+tanx =1−sin2x1+cosxsinx  −cos2x1+sinxcosx    (Đk: sinx và cosx khác 0)

=1−sin3xsinx+cosx −cos3xcosx+sinx 

=1−(sinx+cosx)(sin2x−sinx.cosx+cos2x)sinx+cosx 

=1−(sin2x+cos2x−sinx.cosx) (do sinx + cosx luôn khác 0)

=sinx.cosx ( do sin2x+cos2x=1)

b) Ta có: sin2x+2cosx−12+cosx−cos2x =(sin2x−1)+2cosx−(cosx+1)(cosx−2)  (Đk: cosx khác -1 và 2)

=−cosx(cosx−2)−(cosx+1)(cosx−2) 

=cosx1+cosx 

ăerw

a.

Đặt \(y=\dfrac{2sinx+cosx}{sinx-cosx+3}\)

\(\Leftrightarrow y.sinx-y.cosx+3y=2sinx+cosx\)

\(\Leftrightarrow\left(2-y\right)sinx+\left(y+1\right)cosx=3y\)

Theo điều kiện có nghiệm của pt lượng giác bậc nhất:

\(\left(2-y\right)^2+\left(y+1\right)^2\ge9y^2\)

\(\Leftrightarrow7y^2+2y-5\le0\)

\(\Leftrightarrow-1\le y\le\dfrac{5}{7}\) (đpcm)

b.

Hoàn toàn tương tự câu a:

Đặt \(y=\dfrac{2sinx+cosx+2}{2cosx-sinx+4}\)

\(\Leftrightarrow2y.cosx-y.sinx+4y=2sinx+cosx+2\)

\(\Leftrightarrow\left(y+2\right)sinx+\left(1-2y\right)cosx=4y-2\)

Theo đk có nghiệm pt lượng giác bậc nhất:

\(\left(y+2\right)^2+\left(1-2y\right)^2\ge\left(4y-2\right)^2\)

\(\Leftrightarrow11y^2-16y-1\le0\)

\(\Leftrightarrow\dfrac{8-5\sqrt{3}}{11}\le y\le\dfrac{8+5\sqrt{3}}{11}\)

Đề bài chắc sai, em kiểm tra lại số liệu đề câu b nhé

a+b+c : dựa vào cái hệ thức \(\sin^2\alpha+\cos^2\alpha=1\)

a) Ta có :  \(\left(\sin x+\cos x\right)^2\)

\(=\sin^2x+2.\sin x.\cos x+\cos^2x\)

\(=1+2.\sin x.\cos x\left(đpcm\right)\)

b) Ta có :  \(\left(\sin x+\cos x\right)^2+\left(\sin x-\cos x\right)^2\)

\(=\sin^2x+2.\sin x.\cos x+\cos^2x+\sin^2x-2.\sin x.\cos x+\cos^2x\)

\(=\sin^2x+\cos^2x+\sin^2x+\cos^2x\)

\(=2\left(\sin^2x+\cos^2x\right)\)

\(=2\times1=2\left(đpcm\right)\)

c) Ta có :  \(\sin^4x+\cos^4x\)

\(=\left(\sin^2x\right)^2+\left(\cos^2x\right)^2\)

\(=\left(\sin^2x+\cos^2x\right)^2-2.\sin^2x.\cos^2x\)

\(=1-2.\sin^2x.\cos^2x\left(đpcm\right)\)

Vậy ...

1, \(sin\left(x+\dfrac{\pi}{6}\right)+cos\left(x+\dfrac{\pi}{6}\right)=\dfrac{\sqrt{6}}{2}\)

⇔  \(\dfrac{\sqrt{2}}{2}sin\left(x+\dfrac{\pi}{6}\right)+\dfrac{\sqrt{2}}{2}cos\left(x+\dfrac{\pi}{6}\right)=\dfrac{\sqrt{3}}{2}\)

⇔ \(sin\left(x+\dfrac{\pi}{6}+\dfrac{\pi}{4}\right)=sin\dfrac{\pi}{4}\)

2, \(\left(\sqrt{3}-1\right)sinx+\left(\sqrt{3}+1\right)cosx=1-\sqrt{3}\)

⇔ \(\dfrac{\left(\sqrt{3}-1\right)}{2\sqrt{2}}sinx+\dfrac{\left(\sqrt{3}+1\right)}{2\sqrt{2}}cosx=\dfrac{1-\sqrt{3}}{2\sqrt{2}}\)

⇔ sinx . si