Tolong Kaka Bantu Jawab Yang Pinter Matematika, Ada 4 Soal

Berikut ini adalah pertanyaan dari HanabiYos pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Atas

Tolong Kaka Bantu Jawab Yang Pinter Matematika, Ada 4 Soal Di Gambar Pada Pojok kanan bawah, Terimakasih :))))))((

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

$\purple{\large{1.}}$

$\cos~x=\frac{1}{2}$

$\cos~x=\cos~60\degree$

$a.~x=60\degree$

$x=60\degree+k.\left(360\degree\right)$

$\to k=0~:~x=60\degree+(0).\left(360\degree\right)=60\degree$

$\to k=1~:~x=60\degree+(1).\left(360\degree\right)=420\degree$ $\sf (tidak~$ $\sf berada~$ $\sf pada~$ $\sf interval~$ $0\degree \leqslant x \leqslant 360\degree)$

$b.~x=-60\degree$

$x=-60\degree+k.\left(360\degree\right)$

$\to k=0~:~x=-60\degree+(0).\left(360\degree\right)=-60\degree$ $\sf (tidak~$ $\sf berada~$ $\sf pada~$ $\sf interval~$ $0\degree \leqslant x \leqslant 360\degree)$

$\to k=1~:~x=-60\degree+(1).\left(360\degree\right)=300\degree$

$\to k=2~:~x=-60\degree+(2).\left(360\degree\right)=660\degree$ $\sf (tidak~$ $\sf berada~$ $\sf pada~$ $\sf interval~$ $0\degree \leqslant x \leqslant 360\degree)$

Jadi : $\huge{\red{\sf HP=\left\{~60\degree,300\degree~\right\}}}$

$\\$

$\purple{\large{2.}}$

$\cos~\left(x-1\degree\right)=\frac{1}{2}\sqrt{2}$

$\cos~\left(x-1\degree\right)=\cos~45\degree$

$a.~x-1\degree=45\degree$

$x-1\degree=45\degree+k.\left(360\degree\right)$

$x=45\degree+1\degree+k.\left(360\degree\right)$

$x=46\degree+k.\left(360\degree\right)$

$\to k=0~:~x=46\degree+(0).\left(360\degree\right)=46\degree$

$\to k=1~:~x=46\degree+(1).\left(360\degree\right)=406\degree$ $\sf (tidak~$ $\sf berada~$ $\sf pada~$ $\sf interval~$ $0\degree \leqslant x \leqslant 360\degree$

$b.~x-1\degree=-45\degree$

$x=-45\degree+1\degree+k.\left(360\degree\right)$

$x=-44\degree+k.\left(360\degree\right)$

$\to k=0~:~x=-44\degree+(0).\left(360\degree\right)=-44\degree$ $\sf (tidak~$ $\sf berada~$ $\sf pada~$ $\sf interval~$ $0\degree \leqslant x \leqslant 360\degree)$

$\to k=1~:~x=-44\degree+(1).\left(360\degree\right)=316\degree$

$\to k=2~:~x=-44\degree+(2).\left(360\degree\right)=676\degree$ $\sf (tidak~$ $\sf berada~$ $\sf pada~$ $\sf interval~$ $0\degree \leqslant x \leqslant 360\degree$

Jadi : $\huge{\red{\sf HP=\left\{~46\degree,316\degree~\right\}}}$

$\\$

$\purple{\large{3.}}$

$\sin~\left(x+8\degree\right)=\frac{1}{2}\sqrt{3}$

$\sin~\left(x+8\degree\right)=\sin~60\degree$

$a.~x+8\degree=60\degree$

$x+8\degree=60\degree+k.\left(360\degree\right)$

$x=60\degree-8\degree+k.\left(360\degree\right)$

$x=52\degree+k.\left(360\degree\right)$

$\to k=0~:~x=52\degree+(0).\left(360\degree\right)=52\degree$

$\to k=1~:~x=52\degree+(1).\left(360\degree\right)=412\degree$ $\sf (tidak~$ $\sf berada~$ $\sf pada~$ $\sf interval~$ $0\degree \leqslant x \leqslant 360\degree)$

$b.~x+8\degree=180\degree-60\degree$

$x+8\degree=120\degree$

$x+8\degree=120\degree+k.\left(360\degree\right)$

$x=120\degree-8\degree+k.\left(360\degree\right)$

$x=112\degree+k.\left(360\degree\right)$

$\to k=0~:~x=112\degree+(0).\left(360\degree\right)=112\degree$

$\to k=1~:~x=112\degree+(1).\left(360\degree\right)=472\degree$ $\sf (tidak~$ $\sf berada~$ $\sf pada~$ $\sf interval~$ $0\degree \leqslant x \leqslant 360\degree)$

Jadi : $\huge{\red{\sf HP=\left\{~52\degree,112\degree~\right\}}}$

$\\$

$\purple{\large{4.}}$

$\tan~\left(x-50\degree\right)=\frac{1}{3}\sqrt{3}$

$\tan~\left(x-50\degree\right)=\tan~30\degree$

$a.~x-50\degree=30\degree$

$x-50\degree=30\degree+k.\left(360\degree\right)$

$x=30\degree+50\degree+k.\left(360\degree\right)$

$x=80\degree+k.\left(360\degree\right)$

$\to k=0~:~x=80\degree+(0).\left(360\degree\right)=80\degree$

$\to k=1~:~x=80\degree+(1).\left(360\degree\right)=440\degree$ $\sf (tidak~$ $\sf berada~$ $\sf pada~$ $\sf interval~$ $0\degree \leqslant x \leqslant 360\degree)$

$b.~x-50\degree=180\degree+30\degree$

$x-50\degree=210\degree$

$x-50\degree=210\degree+k.\left(360\degree\right)$

$x=210\degree+50\degree+k.\left(360\degree\right)$

$x=260\degree+k.\left(360\degree\right)$

$\to k=0~:~x=260\degree+(0).\left(360\degree\right)=260\degree$

$\to k=1~:~x=260\degree+(1).\left(360\degree\right)=620\degree$ $\sf (tidak~$ $\sf berada~$ $\sf pada~$ $\sf interval~$ $0\degree \leqslant x \leqslant 360\degree)$

Jadi : $\huge{\red{\sf HP=\left\{~80\degree,260\degree~\right\}}}$

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Last Update: Sun, 25 Jul 21

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