6. `cos^2 2x=1/4`

`<=>cos^2 2x = (1/2)^2`

TH1: `cos 2x=1/2`

`<=>cos2x=cos pi/3`

`<=>2x=+- pi/3+k2pi`

`<=>x=+-pi/6+kpi`

``

TH2: `cos2x=-1/2`

`<=>cos2x=cos(2pi)/3`

`<=>2x=+-2pi/3+k2pi`

`<=>x=+-pi/3+kpi`

``

Vậy PT có 2 họ nghiệm như trên.

``

7. `sin^2x=1/2`

`<=>sin^2x=(sqrt2/2)^2`

TH1: `sin^2x=sqrt2/2`

`<=>sinx=sinpi/4`

`<=>[(x=pi/4+k2pi),(x=pi-pi/4+k2pi):}`

`<=>[(x=pi/4+k2pi),(x=(3pi)/4+k2pi):}`

``

TH2: `sinx=-sqrt2/2`

`<=>sinx=sin(-pi/4)`

`<=>[(x=-pi/4+k2pi),(x=pi+pi/4+k2pi):}`

`<=> [( x= -pi/4+k2pi),(x=(5pi)/4 +k2pi):}`

``

Vậy PT có 4 họ nghiệm như trên.

``

8. ĐK: `sin (x+pi/5) ne 0 <=> x +pi/5 ne kpi <=> x ne -pi/5 + kpi`

`3cot^2 (x+pi/5)=1`

`<=>cot^2 (x+pi/5) =1/3`

`<=> cot^2 (x+pi/5) =(sqrt3/3)^2`

`<=>cot (x+pi/5)=+-sqrt3/3`

`<=>cot(x+pi/5)=+-cot(pi/3)`

`<=> [(x+pi/5=pi/3+kpi),(x+pi/5=-pi/3+kpi):}`

`<=>[(x=(2pi)/15+kpi),(x=(-8pi)/15+kpi):}`

Vậy PT có 2 họ nghiệm như trên.

``

9. `|sin(x+pi/6)+1/2|=1/2`

TH1: `sin(x+pi/6)+1/2=1/2`

`<=>sin(x+pi/6)=0`

`<=>x+pi/6=kpi`

`<=>x=-pi/6+kpi` (TM)

TH2: `sin(x+pi/6)+1/2=-1/2`

`<=>sin(x+pi/6)=-1`

`<=>x+pi/6=-pi/2+k2pi`

`<=>x=-(-2pi)/3+k2pi` (TM)

``

Vậy phương trình có 2 họ nghiệm như trên.

``

10. `sin^2 2x+cos^2 3x=1`

`<=>sin^2 2x +cos^2 3x= sin^2 3x + cos^2 3x`

`<=>sin^2 2x=sin^2 3x`

`<=>(sin2x+sin3x)(sin2x-sin3x)=0`

`<=>[(sin2x=-sin3x),(sin2x=sin3x):}`

`<=>[(sin2x=sin(-3x)),(sin2x=sin3x):}`

`<=>[(2x=-3x+k2pi),(2x=pi+3x+k2pi),(2x=3x+k2pi),(2x=pi-3x+k2pi):}`

`<=>[(x=(k2pi)/5),(x=-pi+k2pi),(x=k2pi),(x=pi/5+(k2pi)/5):}`

``

Vậy PT có 4 họ nghiệm như trên.

``

`6)`

`cos^2 2x=1/4`

`<=>`\(\left[ \begin{array}{l}\cos 2x=\dfrac{1}{2}\\\cos 2x=-\dfrac{1}{2}\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}\cos 2x=\cos \dfrac{\pi}{3}\\\cos 2x=-\cos \dfrac{\pi}{3}\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}2x=\dfrac{\pi}{3}+k2\pi\\2x=-\dfrac{\pi}{3}+k2\pi\\2x=\dfrac{2\pi}{3}+k2\pi\\2x=-\dfrac{2\pi}{3}+k2\pi\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=±\dfrac{\pi}{6}+k\pi\\x=±\dfrac{\pi}{3}+k\pi\end{array}\quad (k \in \Bbb Z) \right.\) 

`7)`

`sin^2 x=1/2`

`<=>`\(\left[ \begin{array}{l}\sin x=\dfrac{\sqrt{2}}{2}\\\sin x=-\dfrac{\sqrt2}{2}\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}\sin x=\sin \dfrac{\pi}{4}\\\sin x=\sin \bigg(-\dfrac{\pi}{4}\bigg)\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=\dfrac{\pi}{4}+k2\pi\\x=\pi-\dfrac{\pi}{4}+k2\pi\\x=-\dfrac{\pi}{4}+k2\pi\\x=\pi+\dfrac{\pi}{4}+k2\pi\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=\dfrac{\pi}{4}+k2\pi\\x=\dfrac{3\pi}{4}+k2\pi\\x=-\dfrac{\pi}{4}+k2\pi\\x=\dfrac{5\pi}{4}+k2\pi\end{array}\quad (k \in \Bbb Z)  \right.\) 

`8)`

ĐKXĐ: `sin (x+pi/5) \ne 0<=> x+ pi/5 \ne kpi<=>x \ne -pi/5+kpi, k in ZZ`

`3cot^2 (x+pi/5)=1`

`<=> cot^2 (x+pi/5)=1/3`

`<=>`\(\left[ \begin{array}{l}\cot \bigg(x+\dfrac{\pi}{5}\bigg)=\dfrac{\sqrt{3}}{3}\\\cot \bigg(x+\dfrac{\pi}{5}\bigg)=-\dfrac{\sqrt{3}}{3}\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}\cot\bigg(x+\dfrac{\pi}{5}\bigg)=\cot \dfrac{\pi}{3}\\\cot\bigg(x+\dfrac{\pi}{5}\bigg)=-\cot \dfrac{\pi}{3}\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x+\dfrac{\pi}{5}=\dfrac{\pi}{3}+k\pi\\x+\dfrac{\pi}{5}=-\dfrac{\pi}{3}+k\pi\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=\dfrac{2\pi}{15}+k\pi\\x=-\dfrac{8\pi}{15}+k\pi\end{array}\quad (k \in \Bbb Z) \right.\) 

`9)`

`|sin(x+pi/6)+1/2|=1/2`

`<=>`\(\left[ \begin{array}{l}\sin\bigg(x+\dfrac{\pi}{6}\bigg)+\dfrac{1}{2}=\dfrac{1}{2}\\\sin\bigg(x+\dfrac{\pi}{6}\bigg)+\dfrac{1}{2}=-\dfrac{1}{2}\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}\sin\bigg(x+\dfrac{\pi}{6}\bigg)=0\\\sin\bigg(x+\dfrac{\pi}{6}\bigg)=-1\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x+\dfrac{\pi}{6}=k\pi\\x+\dfrac{\pi}{6}=-\dfrac{\pi}{2}+k2\pi\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=-\dfrac{\pi}{6}+k\pi\\x=-\dfrac{2\pi}{3}+k2\pi\end{array}\quad (k \in \Bbb Z) \right.\) 

`10)`

`sin^2 2x+cos^2 3x=1`

`<=> (1-cos 4x)/2+(1+cos 6x)/2=1`

`<=> cos 6x=cos 4x`

`<=>`\(\left[ \begin{array}{l}6x=4x+k2\pi\\6x=-4x+k\pi\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}2x=k2\pi\\10x=k2\pi\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=k\pi\\x=\dfrac{k\pi}{5} \end{array}\quad (k \in \Bbb Z) \right.\)