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

Vì \(90^o< x< 180^o\Rightarrow\cos x< 0\)

Có: \(\sin^2x+\cos^2x=1\Leftrightarrow\left(\dfrac{2}{5}\right)^2+\cos^2x=1\Leftrightarrow\cos^2x=\dfrac{21}{25}\Leftrightarrow\cos x=-\dfrac{\sqrt{21}}{5}\left(vì\cos x< 0\right)\)

a) Dùng bảng lượng giác sinx = 0,2368 => x ≈ 13o42'

- Cách nhấn máy tính:

b) x ≈ 51o31'

- Cách nhấn máy tính:

c) x ≈ 65o6'

- Cách nhấn máy tính:

d) x ≈ 17o6'

- Cách nhấn máy tính:

\(\tan x=\frac{\sin x}{\cos x}=\frac{3}{5}\Rightarrow\sin x=\frac{3}{5}\cos x\)

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

\(=\frac{\frac{3}{5}.\cos^2x}{\left(\frac{3}{5}\cos x-\cos x\right)\left(\frac{3}{5}\cos x+\cos x\right)}=\frac{\frac{3}{5}\cos^2x}{\frac{-16}{25}.\cos^2x}=\frac{-15}{16}\)

a: BH=0,5dm=5cm

ΔAHB vuông tại H

=>AH^2+HB^2=AB^2

=>AH^2=13^2-5^2=12^2

=>AH=12cm

sin B=AH/AB=12/13

sin C=sin HAC=BH/AB=5/13

b: ΔABC vuông tại A có AH là đường cao

nên AH^2=HB*HC

=>AH=2*căn 3(cm)

BC=3+4=7cm

\(AB=\sqrt{BH\cdot BC}=\sqrt{21}\left(cm\right)\)

\(AC=\sqrt{4\cdot7}=2\sqrt{7}\left(cm\right)\)

Xét ΔABC vuông tại A có

sin C=AB/BC=căn 21/7

sin B=AC/BC=2/căn 7

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

\(\Rightarrow sinx=\sqrt{1-cos^2x}=\sqrt{1-\left(-\dfrac{3}{5}\right)^2}=\dfrac{4}{5}\)

\(sin\left(x+\dfrac{\pi}{3}\right)=sinx.cos\left(\dfrac{\pi}{3}\right)+cosx.sin\left(\dfrac{\pi}{3}\right)=\dfrac{4}{5}.\dfrac{1}{2}+\left(-\dfrac{3}{5}\right).\dfrac{\sqrt{3}}{2}=\dfrac{4-3\sqrt{3}}{10}\)

\(cos\left(x+\dfrac{\pi}{4}\right)=cosx.cos\left(\dfrac{\pi}{4}\right)-sinx.sin\left(\dfrac{\pi}{4}\right)=-\dfrac{3}{5}.\dfrac{\sqrt{2}}{2}-\dfrac{4}{5}.\dfrac{\sqrt{2}}{2}=-\dfrac{7\sqrt{2}}{10}\)