\(\Leftrightarrow2sin^3x+1-sin^2x-1=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)

Lời giải:

$m^2=(\sin x+\cos x)^2=\sin ^2x+\cos ^2x+2\sin x\cos x=1+2\sin x\cos x$

$\Rightarrow \sin x\cos x=\frac{m^2-1}{2}$

Ta có:

$|\sin ^3x-\cos ^3x|=|\sin x-\cos x||\sin ^2x+\sin x\cos x+\cos ^2x|$

$=\sqrt{(\sin x-\cos x)^2}|1+\sin x\cos x|$

$=\sqrt{1-2\sin x\cos x}.|1+\sin x\cos x|$

$=\sqrt{1-(m^2-1)}.|1+\frac{m^2-1}{2}|$

$=\sqrt{2-m^2}.\frac{m^2+1}{2}$

\(sinx+cosx=m\\ \Rightarrow sin^2x+cos^2x+2sinx.cosx=m^2\\ \Rightarrow sinx.cosx=\dfrac{1-m^2}{2}\)

Mặt khác: 

\(sinx-cosx=\left(sinx+cosx\right)-2cosx=m-2cosx\)

Có:

\(\left|sin^3x-cos^3x\right|=\left|\left(sinx-cosx\right)\left(sin^2x+sinx.cosx+cos^2x\right)\right|\\ =\left|\left(m-2cosx\right)\left(1+\dfrac{1-m^2}{2}\right)\right|\\ =\left|\left(m-2cosx\right)\left(\dfrac{3-m^2}{2}\right)\right|\)

Botay.com.vn

\(\cos^21^o+\cos^289^o=\cos^21^o+\cos^2\left(90^o-1^o\right)=\cos^21^o+\sin^21^o=1\)

\(\cos^22^o+\cos^288^o=\cos^22^o+\cos^2\left(90^o-2^o\right)=\cos^22^o+\sin^22^o=1\)

.......

\(\cos^244^o+\cos^246^o=\cos^244^o+\cos^2\left(90^o-44^o\right)=\cos^244^o+\sin^244^o=1\)

\(\cos^245^o=\left(\frac{\sqrt{2}}{2}\right)^2=\frac{1}{2}\)

=> \(A=1.44+\frac{1}{2}-\frac{1}{2}=44\)

\(\Leftrightarrow sin8x-\sqrt{2}cos8x=cos6x-\sqrt{2}sin6x\)

\(\Leftrightarrow\dfrac{1}{\sqrt{3}}sin8x-\dfrac{\sqrt{2}}{\sqrt{3}}cos8x=\dfrac{1}{\sqrt{3}}cos6x-\dfrac{\sqrt{2}}{\sqrt{3}}sin6x\)

Đặt \(\dfrac{1}{\sqrt{3}}=cosa\) với \(a\in\left(0;\dfrac{\pi}{2}\right)\Rightarrow\dfrac{\sqrt{2}}{\sqrt{3}}=sina\)

\(\Rightarrow sin8x.cosa-cos8x.sina=cos6x.cosa-sin6x.sina\)

\(\Leftrightarrow sin\left(8x-a\right)=cos\left(6x+a\right)\)

\(\Leftrightarrow sin\left(8x-a\right)=sin\left(\dfrac{\pi}{2}-6x-a\right)\)

\(\Leftrightarrow...\)

Học cái viết đề đi b. Đọc không có ra

đề nè

\(\left(1+cosx\right)\cdot\left(1+4^{cosx}\right)=3\cdot4^{cosx}\)

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

\(\Rightarrow sin^2x+cos^2x+2sinx.cosx=\frac{1}{4}\)

\(\Rightarrow1+2sinx.cosx=\frac{1}{4}\Rightarrow sinx.cosx=-\frac{3}{8}\)

\(T=\frac{1}{sinx}+\frac{1}{cosx}=\frac{sinx+cosx}{sinx.cosx}=\frac{-\frac{1}{4}}{-\frac{3}{8}}=\frac{2}{3}\)

\(1+\tan^2a=1+\frac{\sin^2a}{\cos^2a}=\frac{\cos^2a+\sin^2a}{\cos^2a}=\frac{1}{\cos^2a}\)

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