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a) Ta có: \(\sin^2a^o=\cos^2\left(90^o-a^o\right)\)
Biểu thức trên
\(=\left(\sin^21^o+\sin^o89\right)+\left(\sin^22^o+\sin^288^o\right)+...+\left(\sin^244^o+\sin^246^o\right)+\sin^245^o\)
\(=\left(\sin^21^o+\cos^21^o\right)+\left(\sin^22^o+\cos^22^o\right)+...+\left(\sin^244^o+\cos^246^o\right)+\sin^245^o\)
\(=1+1+..+1+\sin^245^o=44+\frac{1}{2}=\frac{89}{2}\)
b)
Ta có: \(\sin^2x+\cos^2x=1\)
\(0^o< x< 90^o\)
=> \(0< \sin x;\cos x< 1\)
Ta có: \(\frac{\sin^2x+\cos^2x}{\text{}\text{}\sin x.\cos x}=\frac{1}{\frac{12}{25}}=\frac{25}{12}\Leftrightarrow\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}=\frac{25}{12}\)
\(\Leftrightarrow\tan x+\frac{1}{\tan x}=\frac{25}{12}\Leftrightarrow\tan^2x-\frac{25}{12}\tan x+1=0\)
Đặt t =tan x => có phương trình bậc 2 ẩn t => Giải đen ta => ra đc t => ra đc tan t
\(\Leftrightarrow\orbr{\begin{cases}\tan x=\frac{3}{4}\\\tan x=\frac{4}{3}\end{cases}}\)
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) \(cos^2x-3sin^2x=0,19\Leftrightarrow1-sin^2x-3sin^2x=0,19\Leftrightarrow1-4sin^2x=0,19\Leftrightarrow4sin^2x=0,81\Leftrightarrow sin^2x=\dfrac{81}{400}\Leftrightarrow sinx=\dfrac{9}{20}\Leftrightarrow x=26^o44'\)
\(B=\dfrac{1-4\sin^2x\cdot\cos^2x}{\sin^2x+2\sin x\cdot\cos x+\cos^2}+2\sin x\cdot\cos x\\ B=\dfrac{1-4\sin^2x\cdot\cos^2x}{2\sin x\cdot\cos x}+2\sin x\cdot\cos x\\ B=\dfrac{1-4\sin^2x\cdot\cos^2x+4\sin^2x\cdot\cos^2x}{2\sin x\cdot\cos x}=\dfrac{1}{2\sin x\cdot\cos x}\)
c)
\(\cos\left(x\right)^4+\sin\left(x\right)^2\cos\left(x\right)^2+\sin\left(x\right)^2\\ =\left(\cos\left(x\right)^2+\sin\left(x\right)^2\right)\cos\left(x\right)^2+\sin\left(x\right)^2\\ =\cos\left(x\right)^2+\sin\left(x\right)^2\\ =1\)
\(\cos\left(x\right)^4-\sin\left(x\right)^4+2\sin\left(x\right)^2\\ =\left(\cos\left(x\right)^2-\sin\left(x\right)^2\right)\left(\cos\left(x\right)^2+\sin\left(x\right)^2\right)+2\sin\left(x\right)^2\\ =\cos\left(2x\right)\cdot1+2\sin\left(x\right)^2\\ =\cos\left(x\right)^2-\sin\left(x\right)^2+2\sin\left(x\right)^2\\ =\cos\left(x\right)^2+\sin\left(x\right)^2\\ =1\)
1: \(=\dfrac{cotx+1+tanx+1}{\left(tanx+1\right)\left(cotx+1\right)}\)
\(=\dfrac{\dfrac{1}{cotx}+cotx+2}{2+tanx+cotx}\)
\(=1\)
2: \(VT=\dfrac{cos^2x+cosxsinx+sin^2x-sinx\cdot cosx}{sin^2x-cos^2x}\)
\(=\dfrac{1}{sin^2x-cos^2x}\)
\(VP=\dfrac{1+cot^2x}{1-cot^2x}=\left(1+\dfrac{cos^2x}{sin^2x}\right):\left(1-\dfrac{cos^2x}{sin^2x}\right)\)
\(=\dfrac{1}{sin^2x}:\dfrac{sin^2x-cos^2x}{sin^2x}=\dfrac{1}{sin^2x-cos^2x}\)
=>VT=VP
ĐKXĐ: \(cosx\ne\frac{1}{2}\Rightarrow x\ne\pm\frac{\pi}{3}+k2\pi\)
\(cos2x+\sqrt{3}\left(1+sinx\right)=\frac{2cosx-1+4sinx.cosx-2sinx}{2cosx-1}\)
\(\Leftrightarrow cos2x+\sqrt{3}\left(1+sinx\right)=\frac{2cosx-1+2sinx\left(2cosx-1\right)}{2cosx-1}\)
\(\Leftrightarrow cos2x+\sqrt{3}+\sqrt{3}sinx=2sinx+1\)
\(\Leftrightarrow1-2sin^2x+\sqrt{3}\left(1+sinx\right)=2sinx+1\)
\(\Leftrightarrow2sin^2x+2sinx-\sqrt{3}\left(1+sinx\right)=0\)
\(\Leftrightarrow\left(2sinx-\sqrt{3}\right)\left(1+sinx\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}sinx=-1\\sinx=\frac{\sqrt{3}}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{2}+k2\pi\\x=\frac{\pi}{3}+k2\pi\left(ktm\right)\\x=\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
