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

\(\eqalign{ \dfrac{2\tan x}{1-\tan^2x}-5&=0\\ \tan2x-5&=0\\ \tan2x&=5 }\)
\(\eqalign{ mi&=\sqrt{1^2+5^2}\\ &=\sqrt{26} }\)

\(\eqalign{ \cos^2x-\sin^2x&=\cos2x\\ &=\dfrac1{\sqrt{26}} }\)

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}

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

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

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

• 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{cases} \sin(x+y)=1+\dfrac15\cos y\\[8pt] \sin(x-y)=-1+\cos y \end{cases} dengan {0\lt y\lt\dfrac{\pi}2}. Nilai \sin x=

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

4. 1
5. \dfrac56

\(\begin{aligned} \sin(x+y)&=1+\dfrac15\cos y\\[8pt] \sin x\cos y+\cos x\sin y&=1+\dfrac15\cos y \end{aligned}\)

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

\(\begin{aligned} \sin x\cos y+\cos x\sin y&=1+\dfrac15\cos y\\ \sin x\cos y-\cos x\sin y&=-1+\cos y&\qquad\color{red}{+}\\\hline 2\sin x\cos y&=\dfrac65\cos y\\[8pt] 2\sin x&=\dfrac65\\ \sin x&=\boxed{\boxed{\dfrac35}} \end{aligned}\)

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

\(\eqalign{ -2 \csc x + 2 \cot x + 3 \sin x &= 0\\ -\dfrac2{\sin x}+\dfrac{2\cos x}{\sin x}+ 3 \sin x &= 0\qquad&{\color{red}\times\sin x}\\ -2+2\cos x+3\sin^2x&=0\\ -2+2\cos x+3\left(1-\cos^2x\right)&=0\\ -2+2\cos x+3-3\cos^2x&=0\\ -3\cos^2x+2\cos x+1&=0\\ 3\cos^2x-2\cos x-1&=0\\ (3\cos x+1)(\cos x-1)&=0 }\)
\cos x=-\dfrac13 atau \cos x=1

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