\(f\left(-x\right)=\left|-sinx-cosx\right|-\left|-sinx+cosx\right|\)

\(=\left|sinx+cosx\right|-\left|sinx-cosx\right|=-f\left(x\right)\)

\(\Rightarrow f\left(x\right)+f\left(-x\right)=0\)

\(\Rightarrow T=f\left(-\pi\right)+f\left(\pi\right)+f\left(-\frac{\pi}{2}\right)+f\left(\frac{\pi}{2}\right)+...+f\left(-\frac{\pi}{n}\right)+f\left(\frac{\pi}{n}\right)+f\left(0\right)\)

\(=0+0+...+0+f\left(0\right)=f\left(0\right)\)

\(=1-1=0\)

Chứng minh các biểu thức đã cho không phụ thuộc vào x.

Từ đó suy ra f'(x)=0

a) ;

b) ;

c) (\(\sqrt{2}\)-\(\sqrt{6}\))=>f'(x)=0

d,f(x)=\(\frac{3}{2}\)=>f'(x)=0

Hàm số xác định khi: \(\left\{{}\begin{matrix}tanx\ne\pm1;cosx\ne0\\cosx\ne-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\pm\dfrac{\pi}{4}+k\pi\\x\ne\dfrac{\pi}{2}+k\pi\\x\ne\pi+k2\pi\end{matrix}\right.\)

\(f\left(1-x\right)+f\left(x\right)=\dfrac{9^{1-x}}{9^{1-x}+3}+\dfrac{9^x}{9^x+3}=\dfrac{9}{9+3.9^x}+\dfrac{9^x}{9^x+3}=\dfrac{3}{9^x+3}+\dfrac{9^x}{9^x+3}=1\)

\(\Rightarrow f\left(x\right)=1-f\left(1-x\right)\)

\(\Rightarrow f\left(cos^2x\right)=1-f\left(sin^2x\right)\)

Do đó:

\(f\left(3m+\dfrac{1}{4}sinx\right)+f\left(cos^2x\right)=1\)

\(\Leftrightarrow f\left(3m+\dfrac{1}{4}sinx\right)=f\left(sin^2x\right)\) (1)

Hàm \(f\left(x\right)=\dfrac{9^x}{9^x+3}\) có \(f'\left(x\right)=\dfrac{3.9^x.ln9}{\left(9^x+3\right)^2}>0\Rightarrow f\left(x\right)\) đồng biến trên R

\(\Rightarrow\left(1\right)\Leftrightarrow3m+\dfrac{1}{4}sinx=sin^2x\)

Đến đây chắc dễ rồi, biện luận để pt \(sin^2x-\dfrac{1}{4}sinx=3m\) có 8 nghiệm trên khoảng đã cho

\(F\left(x\right)=sin\left(2\pi-\frac{\pi}{2}+x\right)+cos\left(14\pi-\frac{\pi}{2}-x\right)+sin\left(2x+\pi+x\right)-cos\left(6\pi+\pi-x\right)\)

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

\(=-cosx-sinx-sinx+cosx=-2sinx\)

b/ \(F\left(x\right)=-1\Leftrightarrow-2sinx=-1\)

\(\Rightarrow sinx=\frac{1}{2}\Rightarrow x=30^0\)

a) f'(x) = - 3sinx + 4cosx + 5. Do đó

f'(x) = 0 <=> - 3sinx + 4cosx + 5 = 0 <=> 3sinx - 4cosx = 5

<=> sinx - cosx = 1. (1)

Đặt cos φ = , (φ ∈) => sin φ = , ta có:

(1) <=> sinx.cos φ - cosx.sin φ = 1 <=> sin(x - φ) = 1

<=> x - φ = + k2π <=> x = φ + + k2π, k ∈ Z.

b) f'(x) = - cos(π + x) - sin = cosx + sin.

f'(x) = 0 <=> cosx + sin = 0 <=> sin = - cosx <=> sin = sin

<=> = + k2π hoặc = π - x + + k2π

<=> x = π - k4π hoặc x = π + k, (k ∈ Z).

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