a) \(4sinx-1=1\Leftrightarrow4sinx=2\Leftrightarrow sinx=\dfrac{2}{4}=\dfrac{1}{2}\)

\(\Leftrightarrow x=30^o\)

b) \(2\sqrt{3}-3tanx=\sqrt{3}\Leftrightarrow3tanx=2\sqrt{3}-\sqrt{3}=\sqrt{3}\Leftrightarrow tanx=\dfrac{\sqrt{3}}{3}\)

\(\Leftrightarrow x=30^o\)

c) \(7sinx-3cos\left(90^o-x\right)=2,5\Leftrightarrow7sinx-3sinx=2,5\Leftrightarrow4sinx=2,5\Leftrightarrow sinx=\dfrac{5}{8}\Leftrightarrow x=30^o41'\)

d)\(\left(2sin-\sqrt{2}\right)\left(4cos-5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2sin-\sqrt{2}=0\\4cos-5=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}2sin=\sqrt{2}\\4cos=5\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}sin=\dfrac{\sqrt{2}}{2}\\cos=\dfrac{5}{4}\left(loai\right)\end{matrix}\right.\)\(\Rightarrow x=45^o\)

Xin lỗi nãy đang làm thì bấm gửi, quên còn câu e, f nữa:"(

e) \(\dfrac{1}{cos^2x}-tanx=1\Leftrightarrow1+tan^2x-tanx-1=0\Leftrightarrow tan^2x-tanx=0\Leftrightarrow tanx\left(tanx-1\right)=0\Rightarrow tanx-1=0\Leftrightarrow tanx=1\Leftrightarrow x=45^o\)

f) \(cos^2x-3sin^2x=0,19\Leftrightarrow1-sin^2x-3sin^2x=0,19\Leftrightarrow1-4sin^2x=0,19\Leftrightarrow4sin^2x=0,81\Leftrightarrow sin^2x=\dfrac{81}{400}\Leftrightarrow sinx=\dfrac{9}{20}\Leftrightarrow x=26^o44'\)

Đây nè bn:

help me pls T-T

Đề sai nhé, phải là 16,3.

\(7\sin\alpha+13\cos\left(90-\alpha\right)=16,3\)

\(\Leftrightarrow7\sin\alpha+13\sin\alpha=16,3\)

\(\Leftrightarrow20\sin\alpha=16,3\)

\(\Leftrightarrow\sin\alpha=0,815\)

\(\Rightarrow\alpha\approx55\left(độ\right)\)

Viết sai rồi.

sin [a(90-a)]

\(\Rightarrow7sinx-3.cos90.cosx+sin90.sinx=2,5\)

\(\Rightarrow7sinx+sinx=2,5\)

\(\Rightarrow8sinx=2,5\) \(\Rightarrow sinx=\frac{5}{16}\)

\(\left(sin90=1vàcos90=0\right)\)

2.

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

$=1-\frac{1}{2}(2\sin x\cos x)^2=1-\frac{1}{2}\sin ^22x$

Vì: $0\leq \sin ^22x\leq 1$

$\Rightarrow 1\geq 1-\frac{1}{2}\sin ^22x\geq \frac{1}{2}$

Vậy $y_{\max}=1; y_{\min}=\frac{1}{2}$

3.

$0\leq |\sin x|\leq 1$

$\Rightarrow 3\geq 3-2|\sin x|\geq 1$

Vậy $y_{\min}=1; y_{\max}=3$

1.

\(y=\cos x+\cos (x-\frac{\pi}{3})=\cos x+\frac{1}{2}\cos x+\frac{\sqrt{3}}{2}\sin x\)

\(=\frac{3}{2}\cos x+\frac{\sqrt{3}}{2}\sin x\)

\(y^2=(\frac{3}{2}\cos x+\frac{\sqrt{3}}{2}\sin x)^2\leq (\cos ^2x+\sin ^2x)(\frac{9}{4}+\frac{3}{4})\)

\(\Leftrightarrow y^2\leq 3\Rightarrow -\sqrt{3}\leq y\leq \sqrt{3}\)

Vậy $y_{\min}=-\sqrt{3}; y_{max}=\sqrt{3}$

@Nguyễn Việt Lâm giúp em với ạ

a.

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

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

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

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

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

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

b.

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

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

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

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

Xét (1), đặt \(sinx-cosx=t\Rightarrow\left[{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sinx.cosx=\frac{1-t^2}{2}\end{matrix}\right.\)

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

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

\(\Rightarrow\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=1-\sqrt{2}\)

\(\Leftrightarrow sin\left(x-\frac{\pi}{4}\right)=\frac{1-\sqrt{2}}{\sqrt{2}}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+arcsin\left(\frac{1-\sqrt{2}}{\sqrt{2}}\right)+k2\pi\\x=\frac{5\pi}{4}-arcsin\left(\frac{1-\sqrt{2}}{\sqrt{2}}\right)+k2\pi\end{matrix}\right.\)

Lời giải:

$A=\frac{2\cos \frac{2x+y}{2}\sin \frac{x}{2}}{2\sin \frac{2x+y}{2}.\cos \frac{x}{2}}-\frac{2\cos \frac{2x+y}{2}\cos \frac{x}{2}}{-2\sin \frac{2x+y}{2}\sin \frac{x}{2}}$

$=\tan \frac{x}{2}.\cot \frac{2x+y}{2}+\cot \frac{x}{2}.\cot \frac{2x+y}{2}=\cot \frac{2x+y}{2}(\tan \frac{x}{2}+\cot \frac{x}{2})$