\(A=cosa\left(sinb.cosc-cosb.sinc\right)+cosb\left(sinc.cosa-cosc.sina\right)+cosc\left(sinacosb-cosasinb\right)\)

\(A=cosasinbcosc-cosacosbsinc+cosacosbsinc-sinacosbcosc+sinacosbcosc-cosasinbcosc\)

\(A=0\)

\(B=sin^2x+\frac{1}{2}\left(cos\frac{2\pi}{3}+cos2x\right)\)

\(B=\frac{1}{2}-\frac{1}{2}cos2x-\frac{1}{4}+\frac{1}{2}cos2x\)

\(B=\frac{1}{4}\)

\(C=\frac{1}{2}-\frac{1}{2}cos2x+\frac{1}{2}-\frac{1}{2}cos\left(\frac{4\pi}{3}+2x\right)+\frac{1}{2}-\frac{1}{2}cos\left(\frac{4\pi}{3}-2x\right)\)

\(C=\frac{3}{2}-\frac{1}{2}cos2x-\frac{1}{2}\left(cos\left(\frac{4\pi}{3}+2x\right)+cos\left(\frac{4\pi}{3}-2x\right)\right)\)

\(C=\frac{3}{2}-\frac{1}{2}cos2x-cos\frac{4\pi}{3}.cos2x\)

\(C=\frac{3}{2}-\frac{1}{2}cos2x+\frac{1}{2}cos2x\)

\(C=\frac{3}{2}\)

\(D=\frac{1}{2}\left[\sqrt{2}sin\left(\frac{\pi}{4}+x\right)\right]^2-sin^2x-sinx.\sqrt{2}cos\left(\frac{\pi}{4}+x\right)\)

\(D=\frac{1}{2}\left(sinx+cosx\right)^2-sin^2x-sinx\left(sinx-cosx\right)\)

\(D=\frac{1}{2}\left(1+2sinx.cosx\right)-sin^2x-sin^2x+sinx.cosx\)

\(D=\frac{1}{2}+sinxcosx+sinxcosx=\frac{1}{2}+sin2x\)

Góc độ cao của thang dựa vào tường là 60º và chân thang cách tường 4,6 m. Chiều dài của thang là

a/

\(sin^2x-sinx=2\left(1-sin^2x\right)\)

\(\Leftrightarrow3sin^2x-sinx-2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=-1\\sinx=\frac{2}{3}\end{matrix}\right.\)

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

2.

\(2sin^2x+\left(1-\sqrt{3}\right)sinx-\frac{\sqrt{3}}{2}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=-\frac{1}{2}\\sinx=\frac{\sqrt{3}}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\\x=\frac{\pi}{3}+k2\pi\\x=\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)

3.

\(\Leftrightarrow\left[{}\begin{matrix}3x+\frac{\pi}{4}=\frac{\pi}{8}+k2\pi\\3x+\frac{\pi}{4}=-\frac{\pi}{8}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{24}+\frac{k2\pi}{3}\\x=-\frac{\pi}{8}+\frac{k2\pi}{3}\end{matrix}\right.\)

\(1.\sin^2x-\sin x=2\cdot\cos^2x\)

\(\Leftrightarrow\sin^2x-\sin x=2\cdot\left(1-\sin^2x\right)\)

\(\Leftrightarrow3\cdot\sin^2x-\sin x-2=0\)

\(\orbr{\begin{cases}\sin x=1\\\sin x=\frac{-2}{3}\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=\frac{\pi}{2}+k2\pi\\\orbr{\begin{cases}x=arcsin\left(\frac{-2}{3}\right)+k2\pi\\x=\pi-arcsin\left(\frac{-2}{3}\right)+k2\pi\end{cases}}\end{cases}}\)

\(\hept{\begin{cases}x=\frac{\pi}{2}+k2\pi\\x=arcsin\left(\frac{-2}{3}\right)+k2\pi\\x=\pi-arcsin\left(\frac{-2}{3}\right)+k2\pi\end{cases}}\)

Câu 2 bạn coi lại đề

3.

\(1+2sinx.cosx-2cosx+\sqrt{2}sinx+2cosx\left(1-cosx\right)=0\)

\(\Leftrightarrow sin2x-\left(2cos^2x-1\right)+\sqrt{2}sinx=0\)

\(\Leftrightarrow sin2x-cos2x=-\sqrt{2}sinx\)

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

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

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

\(\Leftrightarrow...\)

4.

Bạn coi lại đề, xuất hiện 2 số hạng \(cos4x\) ở vế trái nên chắc là bạn ghi nhầm

5.

\(\Leftrightarrow sinx.sin2x-cosx.sin^22x=2cos^2\left(\frac{\pi}{4}-x\right)-1\)

\(\Leftrightarrow sinx.sin2x-cosx.sin^22x=cos\left(\frac{\pi}{2}-2x\right)\)

\(\Leftrightarrow sinx.sin2x-cosx.sin^22x=sin2x\)

\(\Leftrightarrow sin2x\left(sinx-cosx.sin2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\Leftrightarrow x=...\\sinx-cosx.sin2x-1=0\left(1\right)\end{matrix}\right.\)

Xét (1):

\(\Leftrightarrow sinx-1-2sinx.cos^2x=0\)

\(\Leftrightarrow sinx-1-2sinx\left(1-sin^2x\right)=0\)

\(\Leftrightarrow2sin^3x-sinx-1=0\)

\(\Leftrightarrow\left(sinx-1\right)\left(2sin^2x+2sinx+1\right)=0\)

\(\Leftrightarrow...\)

6.

\(sinx.sin4x=\sqrt{2}cos\left(\frac{\pi}{6}-x\right)-2\sqrt{3}cosx.sin2x.cos2x\)

\(\Leftrightarrow sinx.sin4x=\sqrt{2}cos\left(\frac{\pi}{6}-x\right)-\sqrt{3}cosx.sin4x\)

\(\Leftrightarrow sin4x\left(sinx+\sqrt{3}cosx\right)=\sqrt{2}sin\left(x+\frac{\pi}{3}\right)\)

\(\Leftrightarrow sin4x\left(\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx\right)-\frac{\sqrt{2}}{2}sin\left(x+\frac{\pi}{3}\right)=0\)

\(\Leftrightarrow sin4x.sin\left(x+\frac{\pi}{3}\right)-\frac{\sqrt{2}}{2}sin\left(x+\frac{\pi}{3}\right)=0\)

\(\Leftrightarrow\left(sin4x-\frac{\sqrt{2}}{2}\right)sin\left(x+\frac{\pi}{3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin4x=\frac{\sqrt{2}}{2}\\sin\left(x+\frac{\pi}{3}\right)=0\end{matrix}\right.\)

\(\Leftrightarrow...\)

a. \(sin\left(4x+\pi\right)=sin35^o\)

\(\Leftrightarrow sin\left(4x+180^o\right)=sin35^o\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+180^o=35^o+k.360^o,k\in Z\\4x+180^o=180^o-35^o+k.360^o,k\in Z\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=-145^o+k.360^o,k\in Z\\4x=-35^o+k.360^o,k\in Z\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{145^o}{4}+k.90,k\in Z\\x=-\frac{35^o}{4}+k.90^o,k\in Z\end{matrix}\right.\)

Vậy.....

b.\(sin4x=\frac{1}{5}\)

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

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

Vậy....

c. \(sin\left(x+\frac{8\pi}{7}\right)=3\)

Ta có: \(-1\le sinx\le1\)

\(\Rightarrow-1\le sin\left(3x+\frac{8\pi}{7}\right)\le1\)

Do đó phương trình trên vô nghiệm

d. \(sinx=-7\)

Ta có: \(-1\le sinx\le1\)

Do đó phương trình trên vô nghiệm

e. \(sin\left(3x+\pi\right)=sin\left(2x-3\pi\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+\pi=2x-3\pi+k2\pi,k\in Z\\3x+\pi=\pi-2x+3\pi+k2\pi,k\in Z\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-4\pi+k2\pi,k\in Z\\5x=3\pi+k2\pi,k\in Z\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-4\pi+k2\pi,k\in Z\\x=\frac{3}{5}\pi+\frac{k2\pi}{5},k\in Z\end{matrix}\right.\)

Vậy......

f. \(sin\left(4x-\frac{\pi}{2}\right)=sin\left(\pi-2x\right)\)

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

\(\Leftrightarrow\left[{}\begin{matrix}6x=\frac{3}{2}\pi+k2\pi,k\in Z\\2x=\frac{\pi}{2}+k2\pi,k\in Z\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{3},k\in Z\\x=\frac{\pi}{4}+k\pi,k\in Z\end{matrix}\right.\)

Vậy......

\(A=2cosx-3cosx-sin\left(3\pi+\frac{\pi}{2}-x\right)+tan\left(\pi+\frac{\pi}{2}-x\right)\)

\(A=-cosx+sin\left(\frac{\pi}{2}-x\right)+tan\left(\frac{\pi}{2}-x\right)\)

\(A=-cosx+cosx+cotx=cotx\)

\(B=2cosx+sin\left(4\pi+\pi-x\right)+sin\left(2\pi-\frac{\pi}{2}+x\right)-sinx\)

\(B=2cosx+sin\left(\pi-x\right)+sin\left(-\frac{\pi}{2}+x\right)-sinx\)

\(B=2cosx+sinx-sin\left(\frac{\pi}{2}-x\right)-sinx\)

\(B=2cosx-cosx=cosx\)

\(A=\frac{2sinx.cosx+sinx}{1+2cos^2x-1+cosx}=\frac{sinx\left(2cosx+1\right)}{cosx\left(2cosx+1\right)}=\frac{sinx}{cosx}=tanx\)

\(B=\frac{cosa}{sina}\left(\frac{1+sin^2a}{cosa}-cosa\right)=\frac{cosa}{sina}\left(\frac{1+sin^2a-cos^2a}{cosa}\right)=\frac{cosa}{sina}.\frac{2sin^2a}{cosa}=2sina\)

\(C=\frac{1+cos2x+cosx+cos3x}{2cos^2x-1+cosx}=\frac{1+2cos^2x-1+2cos2x.cosx}{cos2x+cosx}=\frac{2cosx\left(cosx+cos2x\right)}{cos2x+cosx}=2cosx\)

\(D=\frac{2sinx.cosx.\left(-tanx\right)}{-tanx.sinx}-2cosx=2cosx-2cosx=0\)

\(E=cos^2x.cot^2x-cot^2x+cos^2x+2cos^2x+2sin^2x\)

\(E=cot^2x\left(cos^2x-1\right)+cos^2x+2=\frac{cos^2x}{sin^2x}\left(-sin^2x\right)+cos^2x+2=2\)

\(F=\frac{sin^2x\left(1+tan^2x\right)}{cos^2x\left(1+tan^2x\right)}=\frac{sin^2x}{cos^2x}=tan^2x\)

Câu G mẫu số có gì đó sai sai, sao lại là \(2sina-sina?\)

\(H=sin^4\left(\frac{\pi}{2}+a\right)-cos^4\left(\frac{3\pi}{2}-a\right)+1=cos^4a-sin^4a+1\)

\(=\left(cos^2a-sin^2a\right)\left(cos^2a+sin^2a\right)+1=cos^2a-\left(1-cos^2a\right)+1=2cos^2a\)

Bài 2:

\(sin\frac{A+B}{2}=sin\left(\frac{180^0-C}{2}\right)=sin\left(90^0-\frac{C}{2}\right)=cos\frac{C}{2}\)

b/

\(A=cosx+cos\left(x+\frac{2\pi}{3}\right)+cos\left(x+\frac{4\pi}{3}\right)=cosx+2cos\left(x+\pi\right).cos\frac{\pi}{3}\)

\(=cosx-2cosx.\frac{1}{2}=0\)

c/

\(sinA+sinB+sinC=2sin\frac{A+B}{2}cos\frac{A-B}{2}+2sin\frac{C}{2}cos\frac{C}{2}=2cos\frac{C}{2}.cos\frac{A-B}{2}+2sin\frac{C}{2}cos\frac{C}{2}\)

\(=2cos\frac{C}{2}\left(cos\frac{A-B}{2}+sin\frac{C}{2}\right)=2cos\frac{C}{2}\left(cos\frac{A-B}{2}+cos\frac{A+B}{2}\right)=4cos\frac{A}{2}cos\frac{B}{2}cos\frac{C}{2}\)

d/ \(\frac{cos2a}{1+sin2a}=\frac{cos^2a-sin^2a}{cos^2a+sin^2a+2sina.cosa}=\frac{\left(cosa-sina\right)\left(cosa+sina\right)}{\left(cosa+sina\right)^2}=\frac{cosa-sina}{cosa+sina}\)

e/

\(E=\frac{sina+cosa}{cos^3a}=\frac{1}{cos^2a}\left(tana+1\right)=\left(1+tan^2a\right)\left(tana+1\right)\)

\(E=tan^3a+tan^2a+tana+1\)

a.

\(sin\left(2x+1\right)=-cos\left(3x-1\right)\)

\(\Leftrightarrow sin\left(2x+1\right)=sin\left(3x-1-\frac{\pi}{2}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1-\frac{\pi}{2}=2x+1+k2\pi\\3x-1-\frac{\pi}{2}=\pi-2x-1+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+2+k2\pi\\x=\frac{3\pi}{10}+\frac{k2\pi}{5}\end{matrix}\right.\)

b.

\(sin\left(2x-\frac{\pi}{6}\right)=sin\left(\frac{\pi}{4}-x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{6}=\frac{\pi}{4}-x+k2\pi\\2x-\frac{\pi}{6}=\frac{3\pi}{4}+x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5\pi}{36}+\frac{k2\pi}{3}\\x=\frac{11\pi}{12}+k2\pi\end{matrix}\right.\)

c.

\(\Leftrightarrow sin\left(3x+\frac{2\pi}{3}\right)=-sin\left(x-\frac{2\pi}{5}-\pi\right)\)

\(\Leftrightarrow sin\left(3x+\frac{2\pi}{3}\right)=sin\left(x-\frac{2\pi}{5}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+\frac{2\pi}{3}=x-\frac{2\pi}{5}+k2\pi\\3x+\frac{2\pi}{3}=\frac{7\pi}{5}-x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{8\pi}{15}+k\pi\\x=\frac{11\pi}{60}+\frac{k\pi}{2}\end{matrix}\right.\)

d.

\(\Leftrightarrow cos\left(4x+\frac{\pi}{3}\right)=sin\left(\frac{\pi}{4}-x\right)\)

\(\Leftrightarrow cos\left(4x+\frac{\pi}{3}\right)=cos\left(\frac{\pi}{4}+x\right)\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{36}+\frac{k2\pi}{3}\\x=-\frac{7\pi}{60}+\frac{k2\pi}{5}\end{matrix}\right.\)

\(cos\left(2x+\frac{\pi}{6}\right)cos\left(2x-\frac{\pi}{6}\right)=\frac{1}{2}\left(cos4x+cos\frac{\pi}{3}\right)=\frac{1}{2}\left(cos4x+\frac{1}{2}\right)\)

\(sin\left(x+\frac{\pi}{6}\right)sin\left(x-\frac{\pi}{6}\right)=\frac{1}{2}\left(cos\frac{\pi}{3}-cos2x\right)=\frac{1}{2}\left(\frac{1}{2}-cos2x\right)\)

\(\Rightarrow C=\frac{1}{2}sinx.cos4x+\frac{1}{4}sinx+\frac{1}{4}sin3x-\frac{1}{2}sin3x.cos2x\)

\(=\frac{1}{4}sin5x-\frac{1}{4}sin3x+\frac{1}{4}sinx+\frac{1}{4}sin3x-\frac{1}{4}sin5x+\frac{1}{4}sinx\)

\(=\frac{1}{2}sinx\)

a)      

\(\begin{array}{l}\sin \left( {2x - \frac{\pi }{6}} \right) =  - \frac{{\sqrt 3 }}{2}\\ \Leftrightarrow \sin \left( {2x - \frac{\pi }{6}} \right) = \sin \left( { - \frac{\pi }{3}} \right)\end{array}\)

\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}2x - \frac{\pi }{6} =  - \frac{\pi }{3} + k2\pi \\2x - \frac{\pi }{6} = \pi  + \frac{\pi }{3} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}2x =  - \frac{\pi }{6} + k2\pi \\2x = \frac{{3\pi }}{2} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x =  - \frac{\pi }{{12}} + k\pi \\x = \frac{{3\pi }}{4} + k\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)

b)     \(\begin{array}{l}\cos \left( {\frac{{3x}}{2} + \frac{\pi }{4}} \right) = \frac{1}{2}\\ \Leftrightarrow \cos \left( {\frac{{3x}}{2} + \frac{\pi }{4}} \right) = \cos \frac{\pi }{3}\end{array}\)

\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}\frac{{3x}}{2} + \frac{\pi }{4} = \frac{\pi }{3} + k2\pi \\\frac{{3x}}{2} + \frac{\pi }{4} = \frac{{ - \pi }}{3} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{18}} + \frac{{k4\pi }}{3}\\x = \frac{{ - 7\pi }}{{18}} + \frac{{k4\pi }}{3}\end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)

c)       

\(\begin{array}{l}\sin 3x - \cos 5x = 0\\ \Leftrightarrow \sin 3x = \cos 5x\\ \Leftrightarrow \cos 5x = \cos \left( {\frac{\pi }{2} - 3x} \right)\\ \Leftrightarrow \left[ \begin{array}{l}5x = \frac{\pi }{2} - 3x + k2\pi \\5x =  - \left( {\frac{\pi }{2} - 3x} \right) + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}8x = \frac{\pi }{2} + k2\pi \\2x =  - \frac{\pi }{2} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{16}} + \frac{{k\pi }}{4}\\x =  - \frac{\pi }{4} + k\pi \end{array} \right.\end{array}\)

d)      

\(\begin{array}{l}{\cos ^2}x = \frac{1}{4}\\ \Leftrightarrow \left[ \begin{array}{l}\cos x = \frac{1}{2}\\\cos x =  - \frac{1}{2}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\cos x = \cos \frac{\pi }{3}\\\cos x = \cos \frac{{2\pi }}{3}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\left[ \begin{array}{l}x = \frac{\pi }{3} + k2\pi \\x =  - \frac{\pi }{3} + k2\pi \end{array} \right.\\\left[ \begin{array}{l}x = \frac{{2\pi }}{3} + k2\pi \\x =  - \frac{{2\pi }}{3} + k2\pi \end{array} \right.\end{array} \right.\end{array}\)

e)      

\(\begin{array}{l}\sin x - \sqrt 3 \cos x = 0\\ \Leftrightarrow \frac{1}{2}\sin x - \frac{{\sqrt 3 }}{2}\cos x = 0\\ \Leftrightarrow \cos \frac{\pi }{3}.\sin x - \sin \frac{\pi }{3}.\cos x = 0\\ \Leftrightarrow \sin \left( {x - \frac{\pi }{3}} \right) = 0\\ \Leftrightarrow \sin \left( {x - \frac{\pi }{3}} \right) = \sin 0\\ \Leftrightarrow x - \frac{\pi }{3} = k\pi ;k \in Z\\ \Leftrightarrow x = \frac{\pi }{3} + k\pi ;k \in Z\end{array}\)

f)       

\(\begin{array}{l}\sin x + \cos x = 0\\ \Leftrightarrow \frac{{\sqrt 2 }}{2}\sin x + \frac{{\sqrt 2 }}{2}\cos x = 0\\ \Leftrightarrow \cos \frac{\pi }{4}.\sin x + \sin \frac{\pi }{4}.\cos x = 0\\ \Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = 0\\ \Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = \sin 0\\ \Leftrightarrow x + \frac{\pi }{4} = k\pi ;k \in Z\\ \Leftrightarrow x =  - \frac{\pi }{4} + k\pi ;k \in Z\end{array}\)