1.

\(2sin\left(x+\dfrac{\pi}{6}\right)+sinx+2cosx=3\)

\(\Leftrightarrow\sqrt{3}sinx+cosx+sinx+2cosx=3\)

\(\Leftrightarrow\left(\sqrt{3}+1\right)sinx+3cosx=3\)

\(\Leftrightarrow\sqrt{13+2\sqrt{3}}\left[\dfrac{\sqrt{3}+1}{\sqrt{13+2\sqrt{3}}}sinx+\dfrac{3}{\sqrt{13+2\sqrt{3}}}cosx\right]=3\)

Đặt \(\alpha=arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}\)

\(pt\Leftrightarrow\sqrt{13+2\sqrt{3}}sin\left(x+\alpha\right)=3\)

\(\Leftrightarrow sin\left(x+\alpha\right)=\dfrac{3}{\sqrt{13+2\sqrt{3}}}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\alpha=arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\\x+\alpha=\pi-arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\pi-2arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\end{matrix}\right.\)

Vậy phương trình đã cho có nghiệm:

\(x=k2\pi;x=\pi-2arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\)

2.

\(\left(sin2x+cos2x\right)cosx+2cos2x-sinx=0\)

\(\Leftrightarrow2sinx.cos^2x+cos2x.cosx+2cos2x-sinx=0\)

\(\Leftrightarrow\left(2cos^2x-1\right)sinx+cos2x.cosx+2cos2x=0\)

\(\Leftrightarrow cos2x.sinx+cos2x.cosx+2cos2x=0\)

\(\Leftrightarrow cos2x.\left(sinx+cosx+2\right)=0\)

\(\Leftrightarrow cos2x=0\)

\(\Leftrightarrow2x=\dfrac{\pi}{2}+k\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)

Vậy phương trình đã cho có nghiệm \(x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)

$\sin x=0,6\\\Leftrightarrow \sin^2 x=0,36\\\Rightarrow \cos^2 x=0,64\\\Leftrightarrow \cos x=0,8(x>0)$

1.

\(sin^2x+cos^2x=1\Rightarrow\left(\dfrac{1}{4}\right)^2+cos^2x=1\)

\(\Rightarrow cos^2x=\dfrac{15}{16}\Rightarrow cosx=\dfrac{\sqrt{15}}{4}\)

2.

\(tanx=\dfrac{1}{3}\Rightarrow tan^2x=\dfrac{1}{9}\Rightarrow\dfrac{sin^2x}{cos^2x}=\dfrac{1}{9}\)

\(\Rightarrow\dfrac{sin^2x}{1-sin^2x}=\dfrac{1}{9}\Rightarrow9sin^2x=1-sin^2x\)

\(\Rightarrow sin^2x=\dfrac{1}{10}\Rightarrow sinx=\dfrac{\sqrt{10}}{10}\)

iu a ko 

\(\Leftrightarrow2\left(cos^2x-sin^2x\right)+\left(sinx.cosx-1\right)\left(sinx+cosx\right)=0\)

\(\Leftrightarrow\left(sinx+cosx\right)\left(2\left(cosx-sinx\right)+sinx.cosx-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=0\\2\left(cosx-sinx\right)+sinx.cosx-1=0\end{matrix}\right.\)

TH1: \(sinx+cosx=0\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=0\)

\(\Rightarrow x=-\frac{\pi}{4}+k\pi\)

TH2: \(2\left(cosx-sinx\right)+sinx.cosx-1=0\)

Đặt \(cosx-sinx=-\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=a\) (\(\left|a\right|\le\sqrt{2}\))

\(\Rightarrow a^2=1-2sinx.cosx\Rightarrow sinx.cosx=\frac{1-a^2}{2}\)

\(2a+\frac{1-a^2}{2}-1=0\)

\(\Leftrightarrow a^2-4a+1=0\Rightarrow\left[{}\begin{matrix}a=2+\sqrt{3}\left(l\right)\\a=2-\sqrt{3}\end{matrix}\right.\)

\(\Rightarrow-\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=2-\sqrt{3}\)

\(\Rightarrow sin\left(x+\frac{\pi}{4}\right)=\frac{\sqrt{3}-2}{\sqrt{2}}=sin\alpha\)

\(\Rightarrow...\)

Nghiệm thứ 2 xấu vậy, bạn có ghi đề nhầm chỗ nào ko nhỉ?

1.

\(0< x< \dfrac{\pi}{2}\Rightarrow cosx>0\)

\(\Rightarrow cosx=\sqrt{1-sin^2x}=\dfrac{\sqrt{5}}{3}\)

\(tanx=\dfrac{sinx}{cosx}=\dfrac{2}{\sqrt{5}}\)

\(sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\left(sinx+cosx\right)=\dfrac{\sqrt{10}+2\sqrt{2}}{6}\)

2.

Đề bài thiếu, cos?x

Và x thuộc khoảng nào?

3.

\(x\in\left(0;\dfrac{\pi}{2}\right)\Rightarrow sinx;cosx>0\)

\(\dfrac{1}{cos^2x}=1+tan^2x=5\Rightarrow cos^2x=\dfrac{1}{5}\Rightarrow cosx=\dfrac{\sqrt{5}}{5}\)

\(sinx=cosx.tanx=\dfrac{2\sqrt{5}}{5}\)

4.

\(A=\left(2cos^2x-1\right)-2cos^2x+sinx+1=sinx\)

\(B=\dfrac{cos3x+cosx+cos2x}{cos2x}=\dfrac{2cos2x.cosx+cos2x}{cos2x}=\dfrac{cos2x\left(2cosx+1\right)}{cos2x}=2cosx+1\)

ĐK: \(x\ne\dfrac{\pi}{2}+k\pi\)

Ta có: 

\(\left\{{}\begin{matrix}tanx=3\\sin^2x+cos^2x=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}sinx=3cosx\\9cos^2x+cos^2x=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}sinx=3cosx\\cos^2x=\dfrac{1}{10}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}sinx=3cosx\\cosx=\pm\dfrac{1}{\sqrt{10}}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}sinx=\dfrac{3}{\sqrt{10}}\\cosx=\dfrac{1}{\sqrt{10}}\end{matrix}\right.\\\left\{{}\begin{matrix}sinx=-\dfrac{3}{\sqrt{10}}\\cosx=-\dfrac{1}{\sqrt{10}}\end{matrix}\right.\end{matrix}\right.\)

Ta có \(sinx-cosx=\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)\)

a, Do \(0< x< \dfrac{\pi}{4}\Rightarrow-\dfrac{\pi}{4}< x-\dfrac{\pi}{4}< 0\)

⇒ \(\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)\) < 0

⇒ sinx - cosx < 0

=> sinx < cosx

b, Do \(\dfrac{\pi}{4}< x< \dfrac{\pi}{2}\Rightarrow0< x-\dfrac{\pi}{4}< \dfrac{\pi}{4}\)

⇒ \(\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)\) > 0

⇒ sinx - cosx > 0

=> sinx > cosx