3. assalamu'alaikum... kak bantu matematika trigonometri... no. 23.24,25,26,27,28,29,30.... Question from @riorocker10

October 2018 1 129 Report

3. assalamu'alaikum... kak bantu matematika trigonometri...
no. 23.24,25,26,27,28,29,30

fatihahalhusna Nomor 23 (A)
$\displaystyle \frac{\sqrt{4.(\frac{3}{2}\sec x)^2-9}}{\frac{3}{2}\sec x}=\frac{\sqrt{9\sec^2 x-9}}{\frac{3}{2}\sec x}\\\\~~~~~~~~~~~~~~~~~~~~~~~~=\frac{2\tan x}{\sec x}\\\\~~~~~~~~~~~~~~~~~~~~~~~~=\boxed{2\sin x}$

Nomor 24 (E)
$\sin^2 x+\cot^2 x+\cos^2x=1+\cot^2x=\boxed{\csc^2 x}$

Nomor 25 (B)
$\cos^21+\cos^289+\cos^22+\cos^288+.\cos^244+\cos^246+0,5+0\\=\cos^21+\sin^21+\cos^22+\sin^22+.\cos^244+\sin^244+0,5\\=1+1+1...+0,5\\=\frac{88}{2}+0,5\\=\boxed{44,5}$

Nomor 26 (B)
$\displaystyle |\cos x|\sin(x-\frac{3\pi}{2})=-|\cos x|\sin(\frac{3\pi}{2}-x)\\~~~~~~~~~~~~~~~~~~~~~~~~~=|\cos x|\cos x\\\\~~~~~~~~~~~~~~~~~~~~~~~~~=\boxed{\cos^2 x}$

Nomor 27 (B)
$\displaystyle \frac{1-5\cos x+4\cos^2x+4\sin^2x}{2\sin x(1-\cos x)}=7\\\\\frac{5(1-\cos x)}{2\sin x(1-\cos x)}=7\\\\\sin x=\frac{5}{14}\rightarrow \cos x=\boxed{-\frac{\sqrt{171}}{14}}$

Nomor 28 (A)
$\displaystyle \cos^2x-2\sin x\cos x-3\sin^2 x-2\cos^2x=-2\sin^2 x\\-\cos^2x-3\sin x\cos x=0\\3\sin x\cos x =-\cos^2 x\\\tan x=-\frac{1}{3}\\\\\sin x=\frac{1}{\sqrt{10}}\\\\\text{Pada interval}\ 0 \leq x \leq 180\ \text{ada dua nilai sin x, jadi jumlah nilai sin x}\\\text{yang memenuhi adalah: }\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\boxed{\frac{2}{\sqrt{10}}}$

Nomor 29 (D)
$2\cos x=-(1+\sin x)~~~~~~~~~~~(kuadratkan)\\\\4\cos^2x=\sin^2x+2\sin x+1\\4-4\sin^2x=\sin^2x+2\sin x+1\\5\sin^2 x+2\sin x-3=0\\(\sin x+1)(5\sin x-3)=0\\\sin x=-1\ (\text{Tidak memenuhi karena pd interval }0 \leq x \leq \pi\ \text{sin x positif}\\\\\sin x=\frac{3}{5}\\Maka:\\\boxed{\tan x=-\frac{3}{4}}$

Nomor 30 (E)
$\displaystyle \cos(2x-\frac{\pi}{2})=\sqrt{3}\sin(2x-\frac{\pi}{2})\\\\\tan(2x-\frac{\pi}{2})=\sqrt{3}\\\\2x-\frac{\pi}{2}=\frac{\pi}{6}+k.\pi\\\\x=\frac{2\pi}{6}+k.\frac{\pi}{2}\\\text{nilai x terbesar dipenuhi jika :} k=3 \\Jadi \\\boxed{x=\frac{11\pi}{6}}$

$\boxed{\boxed{\bigstar}}}\boxed{\boxed{\bold{Sampai\ Jumpa\ Lagi}}}\boxed{\boxed{\bigstar}}}$