$cos\, x-tan\, x=-\sqrt{1-a^{2}}+\frac{a}{\sqrt{1-a^{2}}}$$=\frac{-\left(1-a^{2}\right)+a}{\sqrt{1-a^{2}}}$$=\frac{a^{2}+a-1}{\sqrt{1-a^{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 $\left\{ \frac{\pi}{4},\frac{3}{4}\pi\right\} .$

$p-q=cos\, A$ (kuaratkan kedua ruas)

$p^{2}+q^{2}-2pq=cos^{2}A$…..(1)

$\sqrt{2pq}=sin\, A$ (kuadratkan kedua ruas)

$2pq=sin^{2}A$….(2)

pers(1) dijumlahkan dengan persamaan (2) diperoleh :

$p^{2}+q^{2}=sin^{2}A+cos^{2}A=1$

$4sin^{2}x+4\cdot cos\, x-1=0$

$4(1-cos^{2}x)+4\cdot cos\, x-1=0$

$4-4\cdot cos^{2}x+4cos\, x-1=0$

$4\cdot cos^{2}x+4\cdot cosx-3=0$

$\left(2cos\, x+1\right)\left(2cos\, x-3\right)=0$

$cos\, x=-\frac{1}{2}$$\rightarrow x=\frac{2}{3}\pi$ dan $-\frac{2}{3}\pi$

$cos\, x=\frac{3}{2}\rightarrow x=\oslash$

$4cos^{2}x-4sin\left(\frac{\pi}{2}+x\right)-3=0$

$4cos^{2}x-4cos\, x-3=0$

$\left(2\cdot cos\, x+1\right)\left(2\cdot cos\, x-3\right)=0$

$cos\, x=\frac{3}{2}$ (tidak memenuhi)

$cos\, x=-\frac{1}{2}$

$x=\pm120^{\circ}=\pm\frac{2}{3}\pi$

$sin\, x+cos\, x=\frac{1}{2}$ (dikuadratkan)

$sin^{2}x+cos^{2}x+2sin\, x\cdot cos\, x=\frac{1}{4}$

$1+2sin\, x\cdot cos\, x=\frac{1}{4}$$\rightarrow2\cdot sin\, x\cdot cos\, x=-\frac{3}{4}$$\rightarrow sin\, x\cdot cos\, x=-\frac{3}{8}$

$tan\, x+cot\, x=\frac{sinx}{cosx}+\frac{cosx}{sinx}$$=\frac{sin^{2}x+cos^{2}x}{sinx.cosx}$$=\frac{1}{-\frac{3}{8}}$$=-\frac{8}{3}$

$tan\, x\cdot sin\, x-cos\, x=sin\, x$ (bagi dengan $sin\, x$)

$tan\, x-\frac{cos\, x}{sin\, x}=1$

$tan\, x-\frac{1}{tan\, x}=1$$\rightarrow tan^{2}x-tan\, x-1=0$

Gunakan rumus abc :

$tan_{1,2}=\frac{1\pm\sqrt{1^{2}-4(1)(-1)}}{2(1)}$$=\frac{1\pm\sqrt{5}}{2}$

$2cos^{2}a=1+2sin\,2a$

$2cos^{2}a-1=2sin\,2a$

$cos\,2a=2\cdot sin\,2a$$\rightarrow\frac{sin\,2a}{cos\,2a}=tan\,2a=\frac{1}{2}$

$tan\,2a=\frac{2tan\, a}{1-tan^{2}a}=\frac{1}{2}$

$4\cdot tan\, a=1-tan^{2}a$$\rightarrow tan^{2}a+4\cdot tan\, a-1=0$

Gunakan rumus abc :

$tan\, a_{1,2}=\frac{-4\pm\sqrt{16+4}}{2}$$=\frac{-4\pm2\sqrt{5}}{2}$$=-2\pm\sqrt{5}$

Karena $a$ lancip, maka $tan\, a=-2+\sqrt{5}$

$sin\,\alpha-2cos\,\alpha=0$

$sin\,\alpha=2cos\,\alpha$$\rightarrow tan\,\alpha=2$

Panjang sisi miring pada segitiga dengan sudut $\alpha$ adalah $\sqrt{2^{2}+1^{2}}=\sqrt{5}$

Selanjutnya didapatkan $sin\,\alpha=\frac{2}{\sqrt{5}}$ dan $cos\,\alpha=\frac{1}{\sqrt{5}}$

Jadi $cos\,2\alpha=cos^{2}\alpha-sin^{2}\alpha$$=\left(\frac{1}{\sqrt{5}}\right)^{2}-\left(\frac{2}{\sqrt{5}}\right)^{2}$$=\frac{1}{5}-\frac{4}{5}$$=-\frac{3}{5}$

$A+B+C=360^{\circ}$$\rightarrow\frac{1}{2}(A+B+C)=180^{\circ}$

$\frac{1}{2}\left(B+C\right)=180^{\circ}-\frac{1}{2}A$

$sin\,\frac{1}{2}\left(B+C\right)=sin\,\left(180^{\circ}-\frac{1}{2}A\right)$

$sin\,\frac{1}{2}\left(B+C\right)=sin\,\frac{1}{2}A$

Jadi $\frac{sin\,\frac{1}{2}A}{sin\,\frac{1}{2}(B+C)}=\frac{sin\,\frac{1}{2}A}{sin\,\frac{1}{2}A}=1$

$tan\,2\alpha+\frac{4}{tan\,\alpha}=0$

$\frac{2\cdot tan\,\alpha}{1-tan^{2}\alpha}+\frac{4}{tan\,\alpha}=0$

$\frac{2\cdot tan^{2}\alpha+4(1-tan^{2}\alpha)}{tan\,\alpha(1-tan^{2}\alpha)}=0$

$2\cdot tan^{2}\alpha-4\cdot tan^{2}\alpha+4=0$$\rightarrow tan^{2}\alpha-2tan^{2}\alpha+2=0$

$tan^{2}\alpha=2$$\rightarrow tan\,\alpha=\pm\sqrt{2}$

Panjang sisi miring adalah pada segitiga dengan sudut $\alpha=\sqrt{\left(\sqrt{2}\right)^{2}+1^{2}}=\sqrt{3}$

Jadi $cos\,\alpha=\frac{1}{3}\sqrt{3}.$

$2cos^{4}\alpha=sin^{2}\alpha$

$2(cos^{2}\alpha)^{2}=sin^{2}\alpha$

$2(1-sin{}^{2}\alpha)^{2}=sin^{2}\alpha$

$2\left(1-2sin^{2}\alpha+sin^{4}\alpha\right)=sin^{2}\alpha$

$2sin^{4}\alpha-4sin^{2}\alpha+2-sin^{2}\alpha=0$

$2sin^{4}\alpha-5sin^{2}\alpha+2=0$

$\left(2.sin^{2}\alpha-1\right)\left(sin^{2}\alpha-2\right)=0$

$sin^{2}\alpha=\frac{1}{2}\rightarrow sin\,\alpha=\frac{1}{2}\sqrt{2}$ (kuadran II)

Karena $sin\,\alpha=\frac{1}{2}\sqrt{2}$, maka $\alpha=45^{0}.$

Jadi $tan\,45^{\circ}=1$

$sin^{2}\alpha=2$$\rightarrow sin\,\alpha=-\sqrt{2}$ (tidak memenuhi)

$\sqrt{2pq}=sin\,\alpha$(dikuadratkan)

$2pq=sin^{2}A$….(1)

$p+q=cos\,\alpha$ (dikuadratkan )

$p^{2}+2pq+q^{2}=cos^{2}\alpha$$\rightarrow p^{2}+q^{2}=cos^{2}\alpha-sin^{2}\alpha$…..(2)

$\left(p-q\right)^{2}=p^{2}+q^{2}-2pq$$=cos^{2}\alpha-sin^{2}\alpha-sin^{2}\alpha$$=(1-sin^{2}A)-2sin^{2}A$

$=1-3sin^{2}A$$=1-3\left(\frac{1-cos\,2\alpha}{2}\right)$$=\frac{3}{2}cos\,2\alpha-\frac{1}{2}$$=\frac{1}{2}\left(3cos\,2\alpha-1\right)$

$cos\, x+sin\, x=\frac{\sqrt{6}}{2}$

$k=\sqrt{1^{2}+1^{2}}=\sqrt{2}$

$tan\,\alpha=\frac{1}{1}=1$$\rightarrow\alpha=45^{\circ}=\frac{\pi}{4}$

Bentuk lain dari $cos\, x+sin\, x=\frac{\sqrt{6}}{2}$ :

$\sqrt{2}\cdot cos\,\left(x-\frac{\pi}{4}\right)=\frac{\sqrt{6}}{2}$

$cos\,\left(x-\frac{\pi}{4}\right)=\frac{\sqrt{3}}{2}=cos\,\frac{\pi}{6}$

$x-\frac{\pi}{4}=\pm\frac{\pi}{6}+k\cdot2\pi$

untuk tanda positif :

$x=\frac{\pi}{4}+\frac{\pi}{6}+k\cdot2\pi$$=\frac{5}{12}\pi+k\cdot2\pi$

Jika $k=0,$ maka $x=\frac{5}{12}\pi$

untuk tanda negatif :

$x=\frac{\pi}{4}-\frac{\pi}{6}+k\cdot2\pi$$=\frac{1}{12}\pi+k\cdot2\pi$

Jika $k=0,$ maka $x=\frac{1}{12}\pi$

Jadi solusi untuk $x$ adalah $\left(\frac{1}{12}\pi,\frac{5}{12}\pi\right).$

$2x^{2}+px+q=0$

Jumlah akar akarnya $=sin\, t+cos\, t=-\frac{p}{2}$….(1)

Hail kali akar-akarnya $=sin\, t\cdot cos\, t=\frac{q}{2}$….(2)

$sin\, t-cos\, t=\frac{1}{\sqrt{3}}$ (kuadratkan kedua ruas)

$\left(sint-cost\right)^{2}=\frac{1}{3}$

$sin^{2}t+cos^{2}t-2\cdot sin\, t\cdot cos\, t=\frac{1}{3}$

$1-2\cdot sint\cdot cos\, t=\frac{1}{3}$$\rightarrow2\cdot sin\, t\cdot cos\, t=\frac{2}{3}$$\rightarrow sin\, t\cdot cos\, t=\frac{1}{3}$

Jika persamaan (1) kedua ruas dikuadratkan maka akan diperoleh :

$sin^{2}t+cos^{2}t+2\cdot sin\, t\cdot cos\, t=\frac{p^{2}}{4}$

$1+\frac{2}{3}=\frac{p^{2}}{4}\rightarrow p^{2}=\frac{20}{3}$

Substitusikan $sin\, t\cdot cos\, t=\frac{1}{3}$ ke pers (2) sehingga diperoleh :

$\frac{q}{2}=\frac{1}{3}\rightarrow q=\frac{2}{3}$

Jadi nilai $p^{2}-7q=\frac{20}{3}-\frac{14}{3}=\frac{6}{3}=2.$