$\sin\frac{2}{3}\pi+\sin\frac{7}{3}\pi=\sin120^{\circ}+\sin420^{\circ}$$=\sin\frac{1}{2}\sqrt{3}+\frac{1}{2}\sqrt{3}=\sqrt{3}$

$\frac{tan\,60^{\circ}-tan\,30^{\circ}}{1+tan60^{\circ}.tan30^{\circ}}=\frac{\sqrt{3}-\frac{\sqrt{3}}{3}}{1+\sqrt{3}.\frac{\sqrt{3}}{3}}$$=\frac{\frac{2\sqrt{3}}{3}}{2}=\frac{1}{3}\sqrt{3}$

$tan^{2}45^{\circ}-cos^{2}30^{\circ}=$$\left(1\right)^{2}-\left(\frac{1}{2}\sqrt{3}\right)^{2}$$=1-\frac{3}{4}=\frac{1}{4}$

$\frac{sin\,270^{\circ}\cdot cos\,135^{\circ}\cdot tan\,135^{\circ}}{sin150^{\circ}\cdot cos225^{\circ}}$$=\frac{\left(-1\right).\left(-\frac{1}{2}\sqrt{2}\right).\left(-1\right)}{\left(\frac{1}{2}\right).\left(-\frac{1}{2}\sqrt{2}\right)}$$=2$

$\cos150^{\circ}+\sin45^{\circ}+\frac{1}{2}\cdot\cot\left(-330^{\circ}\right)$

$-\cos30^{\circ}+\sin45^{\circ}-\frac{1}{2}\cdot\frac{1}{\tan330^{\circ}}$

$\frac{-1}{2}\sqrt{3}+\frac{1}{2}\sqrt{2}+\frac{1}{2}\sqrt{3}=\frac{1}{2}\sqrt{2}$

$\sqrt{2}sin\, x-1=0$

$sin\, x=\frac{1}{\sqrt{2}}=\frac{1}{2}\sqrt{2}$

$sin\, x=sin\,45^{\circ}\rightarrow x=45^{\circ}=\frac{\pi}{4}$

$sin\, x=\left(180^{\circ}-45^{\circ}\right)+k\cdot360^{\circ}$

$x=135^{\circ}+k\cdot360^{\circ}$

Jika $k=0$, maka $x=135^{\circ}=\frac{3}{4}\pi$

Jadi nilai $x$ yang memenuhi adalah $\left\{ \frac{\pi}{4},\frac{3}{4}\pi\right\} .$

$\tan\left(-45^{\circ}\right)+\sin120^{\circ}+\cos225^{\circ}-\cos30^{\circ}$

$=-\tan45^{\circ}+\sin(180^{\circ}-60^{\circ})+\cos\left(180+45\right)^{\circ}-\cos30^{\circ}$

$=-\tan45^{\circ}+\sin60^{\circ}-\cos45^{\circ}-\cos30^{\circ}$

$=-1+\frac{1}{2}\sqrt{3}-\frac{1}{2}\sqrt{2}-\frac{1}{2}\sqrt{3}=-1-\frac{1}{2}\sqrt{2}$

$\frac{sin\,270^{\circ}\cdot cos\,135^{\circ}\cdot tan\,135^{\circ}}{sin\,150^{\circ}\cdot cos\,225^{\circ}}$
=$\frac{sin\,\left(180+90\right)^{\circ}\cdot cos\,(180-45)^{\circ}\cdot tan\,(180-45)^{\circ}}{sin\,(180-30)^{\circ}\cdot cos\,(180+45)^{0}}$

$=\frac{-sin\,90^{\circ}.\left(-cos\,45^{\circ}\right)\cdot\left(-tan\,45^{\circ}\right)}{sin\,30^{\circ}.\left(-cos\,45^{\circ}\right)}=\frac{(-1)\cdot(-\frac{1}{2}\sqrt{2})\cdot\left(-1\right)}{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\sqrt{2}\right)}=2$

$\tan\left(90^{\circ}-x\right)=\cot x$

$\tan\left(x-30^{\circ}\right)=\tan(90^{\circ}-x)$

$x-30^{\circ}=90^{\circ}-x$

$2x=120^{\circ}\rightarrow x=60^{\circ}$

Kuadran III berarti = $\left(180+60\right)^{\circ}=240^{\circ}.$

$sin\,150^{\circ}+cos\,510^{\circ}+tan\,4110^{\circ}-tan\,210^{\circ}$

$=sin\,150^{\circ}+cos\,\left(360^{\circ}+150^{\circ}\right)+tan\,\left(11\cdot360^{\circ}+150^{\circ}\right)-tan\,210^{\circ}$

$=sin\,150^{\circ}+cos\,150^{\circ}+tan\,150^{\circ}-tan\,210^{\circ}$

$=\frac{1}{2}+\left(-\frac{1}{2}\sqrt{3}\right)-\frac{1}{3}\sqrt{3}-\left(\frac{1}{3}\sqrt{3}\right)$

$=\frac{1}{2}-\frac{1}{2}\sqrt{3}-\frac{2}{3}\sqrt{3}=\frac{1}{2}-\frac{7}{6}\sqrt{3}$

$\frac{sin\frac{\pi}{2}+cos\left(\frac{\pi}{2}+\alpha\right)}{cos\frac{\pi}{2}+sin\left(\frac{\pi}{2}+\alpha\right)}$=
$\frac{1-sin\alpha}{cos\alpha}.\frac{1+sin\alpha}{1+sin\alpha}$$=\frac{\left(1-sin^{2}\alpha\right)}{cos\alpha\left(1+sin\alpha\right)}$$=\frac{cos^{2}\alpha}{cos\alpha\left(1+sin\alpha\right)}$$=\frac{cos\alpha}{1+sin\alpha}$

$x+y=270^{\circ}$

$x=270^{\circ}-y$

$cos\, x=cos\left(270^{\circ}-y\right)=-\sin\, y$

$cos\, x+sin\, y=-sin\, y+sin\, y=0$

$\frac{x.cosec^{2}30^{\circ}.sec^{2}45^{\circ}}{8.cos^{2}45^{0}.sin^{2}60^{0}}=tan^{2}60^{\circ}-tan^{2}30^{\circ}$

$\frac{x.\frac{1}{sin^{2}30^{\circ}}\cdot\frac{1}{cos^{2}45^{\circ}}}{8.cos^{2}45^{\circ}.sin^{2}60^{\circ}}=tan^{2}60^{\circ}-tan^{2}30^{\circ}$

$\frac{x\cdot\frac{1}{\frac{1}{4}}\cdot\frac{1}{\frac{2}{4}}}{8\cdot\frac{2}{4}\cdot\frac{3}{4}}=3-\frac{1}{3}\rightarrow\frac{8x}{3}=\frac{8}{3}\rightarrow x=1$

tan $x=\frac{1}{3}\sqrt{3}\rightarrow x=30^{0}$

$3cos\, x+cos\left(x+\frac{1}{2}\pi\right)+sin\,\left(\pi-x\right)$$=3cos\, x-sin\, x+sinx$$=3cos\, x=3\cdot cos\,30^{\circ}$$=3\cdot\frac{1}{2}\sqrt{3}=\frac{3}{2}\sqrt{3}$

Karena $\gamma=90^{\circ}$, maka $\alpha+\beta=90^{\circ}$

$\beta=90^{\circ}-\alpha$…..(1)

cos$\beta=cos\left(90^{\circ}-\alpha\right)=sin\alpha$

sin$\frac{1}{2}\left(\beta+\gamma\right)=cos\beta$

sin$\frac{1}{2}\left(\beta+90^{\circ}\right)=sin\alpha$

$\frac{1}{2}\left(\beta+90^{\circ}\right)=\alpha$$\rightarrow\beta+90^{\circ}=2\alpha$

=$90^{\circ}-\alpha+90=2\alpha\rightarrow3\alpha=180^{\circ}\rightarrow\alpha=30^{\circ}$