SBMPTN Zone : Persamaan Trigonometri

Berikut ini adalah kumpulan soal Persamaan Trigonometri tingkat SBMPTN. Jika ada jawaban yang salah, mohon dikoreksi melalui komentar. Terima kasih.

Tipe:

Standar SBMPTN HOTS

No. 1

Jika \dfrac{2\tan x}{1-\tan^2x}-5=0, dengan 0\lt x\lt\dfrac{\pi}2 maka {\cos^2x-\sin^2x=} ....

1. \dfrac1{\sqrt{26}}
2. \dfrac2{\sqrt{26}}
3. \dfrac3{\sqrt{26}}

4. \dfrac4{\sqrt{26}}
5. \dfrac5{\sqrt{26}}

SBMPTN 2017

$\begin{array}{rl}{\displaystyle \frac{2\tan x}{1-{\tan }^{2}x}}-5& =0\\ \tan 2x-5& =0\\ \tan 2x& =5\end{array}$
$\begin{array}{rl}mi& =\sqrt{1^2+5^2}\\ & =\sqrt{26}\end{array}$

$\begin{array}{rl}{\cos }^{2}x-{\sin }^{2}x& =\cos 2x\\ & ={\displaystyle \frac{1}{\sqrt{26}}}\end{array}$

Jadi, {\cos^2x-\sin^2x=\dfrac1{\sqrt{26}}}. JAWAB: A

No. 2

Jika {a\cos x-b\sin x=c}, maka {a\sin x+b\cos x=}

1. {a^2+b^2+c^2}
2. \sqrt{a^2+b^2+c^2}
3. {a^2+b^2-c^2}

4. \sqrt{a^2+b^2-c^2}
5. {a^2-b^2-c^2}

$\begin{array}{rl}{(a\cos x-b\sin x)}^2+{(a\sin x+b\cos x)}^2& =a^2{\cos }^{2}x-2ab\sin x\cos x+b^2{\sin }^{2}x+a^2{\sin }^{2}x+2ab\sin x\cos x+b^2{\cos }^{2}x\\ c^2+{(a\sin x+b\cos x)}^2& =a^2({\cos }^{2}x+{\sin }^{2}x)+b^2({\sin }^{2}x+{\cos }^{2}x)\\ c^2+{(a\sin x+b\cos x)}^2& =a^2\left(1\right)+b^2\left(1\right)\\ c^2+{(a\sin x+b\cos x)}^2& =a^2+b^2\\ {(a\sin x+b\cos x)}^2& =a^2+b^2-c^2\\ a\sin x+b\cos x& =\boxed{{\displaystyle \boxed{{\displaystyle \sqrt{a^2+b^2-c^2}}}}}\end{array}$

Jadi, a\sin x+b\cos x=\sqrt{a^2+b^2-c^2}.
JAWAB: D

No. 3

Jika 2\cos x\sin x+1=2\cos x+\sin x dengan 0\leq x\leq2\pi, jumlah semua nilai x yang memenuhi persamaan tersebut adalah....

1. \dfrac56\pi
2. \dfrac{13}6\pi
3. 2\pi

4. \dfrac52\pi
5. 3\pi

Grup Facebook

$\begin{array}{rl}2\cos x\sin x+1& =2\cos x+\sin x\\ 2\cos x\sin x-2\cos x-\sin x+1& =0\\ (2\cos x-1)(\sin x-1)& =0\end{array}$

• 2\cos x-1=0
  \cos x=\dfrac12
  x=\left\{\dfrac13\pi,\dfrac53\pi\right\}
• \sin x-1=0
  \sin x=1
  x=\left\{\dfrac12\pi\right\}

\dfrac13\pi+\dfrac53\pi+\dfrac12\pi=\boxed{\boxed{\dfrac52\pi}}

Jadi, jumlah semua nilai x yang memenuhi persamaan tersebut adalah \dfrac52\pi.
JAWAB: D

No. 4

Diketahui sistem persamaan $$\{\begin{array}{l}\sin (x+y)=1+{\displaystyle \frac{1}{5}}\cos y\\ \sin (x-y)=-1+\cos y\end{array}$$ dengan {0\lt y\lt\dfrac{\pi}2}. Nilai \sin x=

1. \dfrac25
2. \dfrac35
3. \dfrac45

4. 1
5. \dfrac56

$\begin{array}{rl}\sin (x+y)& =1+{\displaystyle \frac{1}{5}}\cos y\\ \sin x\cos y+\cos x\sin y& =1+{\displaystyle \frac{1}{5}}\cos y\end{array}$

$\begin{array}{rl}\sin (x-y)& =-1+\cos y\\ \sin x\cos y-\cos x\sin y& =-1+\cos y\end{array}$

$\begin{array}{rl}\sin x\cos y+\cos x\sin y& =1+{\displaystyle \frac{1}{5}}\cos y\\ \sin x\cos y-\cos x\sin y& =-1+\cos y& +\\ 2\sin x\cos y& ={\displaystyle \frac{6}{5}}\cos y\\ 2\sin x& ={\displaystyle \frac{6}{5}}\\ \sin x& =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{3}{5}}}}}}\end{array}$

Jadi,

No. 5

Jika x memenuhi {-2 \csc x + 2 \cot x + 3 \sin x = 0} untuk 0 \lt x \lt π, maka \cos x =

1. -\dfrac23
2. -\dfrac13
3. -\dfrac12

4. \dfrac12
5. \dfrac23

SKMP September 2020

$\begin{array}{rl}-2\csc x+2\cot x+3\sin x& =0\\ -{\displaystyle \frac{2}{\sin x}}+{\displaystyle \frac{2\cos x}{\sin x}}+3\sin x& =0& \times \sin x\\ -2+2\cos x+3{\sin }^{2}x& =0\\ -2+2\cos x+3(1-{\cos }^{2}x)& =0\\ -2+2\cos x+3-3{\cos }^{2}x& =0\\ -3{\cos }^{2}x+2\cos x+1& =0\\ 3{\cos }^{2}x-2\cos x-1& =0\\ (3\cos x+1)(\cos x-1)& =0\end{array}$
\cos x=-\dfrac13 atau \cos x=1

Jadi, \cos x =-\dfrac13.
JAWAB: B