24.

\(cos\left(x-\dfrac{\pi}{2}\right)\le1\Rightarrow y\le3.1+1=4\)

\(y_{max}=4\)

26.

\(y=\sqrt{2}cos\left(2x-\dfrac{\pi}{4}\right)\)

Do \(cos\left(2x-\dfrac{\pi}{4}\right)\le1\Rightarrow y\le\sqrt{2}\)

\(y_{max}=\sqrt{2}\)

b.

\(\dfrac{1}{2}sinx+\dfrac{\sqrt{3}}{2}cosx=\dfrac{1}{2}\)

\(\Leftrightarrow cos\left(x-\dfrac{\pi}{6}\right)=\dfrac{1}{2}\)

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

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

Biến đổi :

\(4\sin x+3\cos x=A\left(\sin x+2\cos x\right)+B\left(\cos x-2\sin x\right)=\left(A-2B\right)\sin x+\left(2A+B\right)\cos x\)

Đồng nhất hệ số hai tử số, ta có :

\(\begin{cases}A-2B=4\\2A+B=3\end{cases}\)\(\Leftrightarrow\begin{cases}A=2\\B=-1\end{cases}\)

Khi đó \(f\left(x\right)=\frac{2\left(\left(\sin x+2\cos x\right)\right)-\left(\left(\sin x-2\cos x\right)\right)}{\left(\sin x+2\cos x\right)}=2-\frac{\cos x-2\sin x}{\sin x+2\cos x}\)

Do đó, 

\(F\left(x\right)=\int f\left(x\right)dx=\int\left(2-\frac{\cos x-2\sin x}{\sin x+2\cos x}\right)dx=2\int dx-\int\frac{\left(\cos x-2\sin x\right)dx}{\sin x+2\cos x}=2x-\ln\left|\sin x+2\cos x\right|+C\)

Ta có :

\(f\left(x\right)=\int\frac{dx}{\sqrt{3}\sin x+\cos x}=\frac{1}{2}\int\frac{dx}{\frac{\sqrt{3}}{2}\sin x+\frac{1}{2}\cos x}=\frac{1}{2}\int\frac{dx}{\sin\left(x+\frac{\pi}{6}\right)}\)

\(=\int\frac{dx}{2\tan\left(\frac{x}{2}+\frac{\pi}{12}\right)\cos^2\left(\frac{x}{2}+\frac{\pi}{12}\right)}=\int\frac{dx}{\sin\left(\frac{x}{2}+\frac{\pi}{12}\right)\cos\left(\frac{x}{2}+\frac{\pi}{12}\right)}=\int\frac{d\left(\tan\frac{x}{2}+\frac{\pi}{12}\right)}{\tan\left(\frac{x}{2}+\frac{\pi}{12}\right)}=\ln\left|\tan\left(\frac{x}{2}+\frac{\pi}{12}\right)\right|+C\)

ko biết

\(y=\sqrt{\dfrac{\sqrt{2}sin\left(2x-\dfrac{\pi}{4}\right)+m-1}{2cos^24x+\dfrac{3}{2}cos4x+\dfrac{21}{2}-5m}}\) 

Hàm xác định trên R khi:

TH1: \(\left\{{}\begin{matrix}\sqrt{2}sin\left(2x-\dfrac{\pi}{4}\right)+m-1\ge0\\2cos^24x+\dfrac{3}{2}cos4x+\dfrac{21}{2}-5m>0\end{matrix}\right.\) ;\(\forall x\)

\(\Rightarrow\left\{{}\begin{matrix}-m\le\min\limits_R\left(\sqrt{2}sin\left(2x-\dfrac{\pi}{4}\right)-1\right)=-1-\sqrt{2}\\5m< \min\limits_R\left(2cos^24x+\dfrac{3}{2}cos4x+\dfrac{21}{2}\right)=\dfrac{327}{32}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m\ge1+\sqrt{2}\\m< \dfrac{327}{160}\end{matrix}\right.\) \(\Rightarrow m\in\varnothing\)

Th2: \(\left\{{}\begin{matrix}\sqrt{2}sin\left(2x-\dfrac{\pi}{4}\right)+m-1\le0\\2cos^24x+\dfrac{3}{2}cos4x+\dfrac{21}{2}-5m< 0\end{matrix}\right.\) ;\(\forall x\)

\(\Rightarrow\left\{{}\begin{matrix}m\le\min\limits_R\left(\sqrt{2}sin\left(2x-\dfrac{\pi}{4}\right)-1\right)=-1-\sqrt{2}\\5m>\max\limits_R\left(2cos^24x+\dfrac{3}{2}cos4x+\dfrac{21}{2}\right)=14\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m\le-1-\sqrt{2}\\m>\dfrac{14}{5}\end{matrix}\right.\) \(\Rightarrow m\in\varnothing\)

Biến đổi :

\(4\sin^2x+1=5\sin^2x+\cos^2x=\left(a\sin x+b\cos x\right)\left(\sqrt{3}\sin x+\cos x\right)+c\left(\sin^2x+\cos^2x\right)\)

\(=\left(a\sqrt{3}+c\right)\sin^2x+\left(a+b\sqrt{3}\right)\sin x.\cos x+\left(b+c\right)\cos^2x\)

Đồng nhấtheej số hai tử số 

\(\begin{cases}a\sqrt{3}+c=5\\a+b\sqrt{3}=0\\b+c=1\end{cases}\)

\(\Leftrightarrow\) \(\begin{cases}a=\sqrt{3}\\b=-1\\c=2\end{cases}\)

a/ Hmm, bạn có nhầm lẫn chỗ nào ko nhỉ, nghiệm của pt này xấu khủng khiếp

b/ \(\Leftrightarrow sin\frac{5x}{2}-cos\frac{5x}{2}-sin\frac{x}{2}-cos\frac{x}{2}=cos\frac{3x}{2}\)

\(\Leftrightarrow2cos\frac{3x}{2}.sinx-2cos\frac{3x}{2}cosx=cos\frac{3x}{2}\)

\(\Leftrightarrow cos\frac{3x}{2}\left(2sinx-2cosx-1\right)=0\)

\(\Leftrightarrow cos\frac{3x}{2}\left(\sqrt{2}sin\left(x-\frac{\pi}{4}\right)-1\right)=0\)

c/ Do \(cosx\ne0\), chia 2 vế cho cosx ta được:

\(3\sqrt{tanx+1}\left(tanx+2\right)=5\left(tanx+3\right)\)

Đặt \(\sqrt{tanx+1}=t\ge0\)

\(\Leftrightarrow3t\left(t^2+1\right)=5\left(t^2+2\right)\)

\(\Leftrightarrow3t^3-5t^2+3t-10=0\)

\(\Leftrightarrow\left(t-2\right)\left(3t^2+t+5\right)=0\)

d/ \(\Leftrightarrow\sqrt{2}\left(\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx\right)=\frac{\sqrt{3}}{2}cos2x-\frac{1}{2}sin2x\)

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

Đặt \(x+\frac{\pi}{3}=a\Rightarrow2x=2a-\frac{2\pi}{3}\Rightarrow2x-\frac{\pi}{3}=2a-\pi\)

\(\sqrt{2}sina=-sin\left(2a-\pi\right)=sin2a=2sina.cosa\)

\(\Leftrightarrow\sqrt{2}sina\left(\sqrt{2}cosa-1\right)=0\)