🌼𝗸𝘂𝗺𝗽𝘂𝗹𝗮𝗻 𝘀𝗼𝗮𝗹 𝗽𝗲𝗿𝘀𝗮𝗺𝗮𝗮𝗻 𝘁𝗿𝗶𝗴𝗼𝗻𝗼𝗺𝗲𝘁𝗿𝗶 𝗸𝗲𝗹𝗼𝗺𝗽𝗼𝗸 𝟭🌼

𝟏) 𝐓𝐞𝐧𝐭𝐮𝐤𝐚𝐧 𝐇𝐢𝐦𝐩𝐮𝐧𝐚𝐧 𝐩𝐞𝐫𝐬𝐚𝐦𝐚𝐚𝐧 𝐝𝐚𝐫𝐢 𝐓𝐚𝐧(𝟐𝐱-𝟏𝟓°)=𝟏, 𝟎° ≤ 𝐱 ≤ 𝟑𝟔𝟎°

𝐉𝐀𝐖𝐀𝐁:

𝐭𝐚𝐧 𝟏 = 𝐓𝐚𝐧 𝟒𝟓° 

=> 𝟐𝐱-𝟏𝟓° = 𝟒𝟓+𝟏×𝟏𝟖𝟎°

𝟐𝐱-𝟏𝟓° = 𝟒𝟓+𝟏𝟖𝟎

𝟐𝐱 = 𝟒𝟓+𝟏𝟖𝟎+𝟏𝟓

𝐱 = 𝟐𝟒𝟎/𝟐 = 𝟏𝟐𝟎

 

=> 𝟐𝐱-𝟏𝟓 = 𝟒𝟓+𝟐×𝟏𝟖𝟎°

𝟐𝐱-𝟏𝟓 = 𝟒𝟓+𝟑𝟔𝟎

𝟐𝐱= 𝟒𝟓+𝟑𝟔𝟎+𝟏𝟓

𝐱 = 𝟒𝟐𝟎/𝟐 = 𝟐𝟏𝟎

 

=> 𝟐𝐱-𝟏𝟓 = 𝟒𝟓+𝟑×𝟏𝟖𝟎

𝟐𝐱-𝟏𝟓 = 𝟒𝟓+𝟓𝟒𝟎

𝟐𝐱 = 𝟒𝟓+𝟓𝟒𝟎+𝟏𝟓

𝐱 = 𝟔𝟎𝟎/𝟐 = 𝟑𝟎𝟎

 

=> 𝟐𝐱-𝟏𝟓 = 𝟒𝟓+𝟒×𝟏𝟖𝟎

𝟐𝐱-𝟏𝟓 = 𝟒𝟓+𝟕𝟐𝟎

𝟐𝐱 = 𝟒𝟓+𝟕𝟐𝟎+𝟏𝟓

𝐱 = 𝟕𝟖𝟎/𝟐 = 𝟑𝟗𝟎° (𝐓𝐢𝐝𝐚𝐤 𝐦𝐞𝐦𝐞𝐧𝐮𝐡𝐢 𝐤𝐚𝐫𝐞𝐧𝐚 𝟎°≤𝐱≤𝟑𝟔𝟎°)

𝐇𝐏 𝐧𝐲𝐚 𝐚𝐝𝐚𝐥𝐚𝐡 {𝟏𝟐𝟎°, 𝟐𝟏𝟎°, 𝟑𝟎𝟎°}

𝟐) 𝐓𝐞𝐧𝐭𝐮𝐤𝐚𝐧 𝐩𝐞𝐧𝐲𝐞𝐥𝐞𝐬𝐚𝐢𝐚𝐧 𝐝𝐚𝐫𝐢 

𝐒𝐢𝐧 𝐱 = ½ √𝟑 𝐮𝐧𝐭𝐮𝐤 𝟎° ≤ 𝐱 ≤ 𝟑𝟔𝟎°

𝐉𝐚𝐰𝐚𝐛:

𝐒𝐢𝐧 𝐱 = 𝐒𝐢𝐧 𝟔𝟎°

𝐱 = 𝟔𝟎° + 𝐤 . 𝟑𝟔𝟎°

𝐤 = 𝟎 —> 𝐱 = 𝟔𝟎°

𝐤 = 𝟏 —> 𝐱 = 𝟒𝟐𝟎° (𝐓𝐌)

𝐀𝐭𝐚𝐮

𝐱 = (𝟏𝟖𝟎° – 𝟔𝟎°) + 𝐤 . 𝟑𝟔𝟎°

𝐱 = 𝟏𝟐𝟎° + 𝐤 . 𝟑𝟔𝟎°

𝐤 = 𝟎 —> 𝐱 = 𝟏𝟐𝟎°

𝐤 = 𝟏 —> 𝐱 = 𝟒𝟖𝟎° (𝐓𝐌)

𝐉𝐚𝐝𝐢, 𝐡𝐢𝐦𝐩𝐮𝐧𝐚𝐧 𝐩𝐞𝐧𝐲𝐞𝐥𝐞𝐬𝐚𝐢𝐚𝐧𝐧𝐲𝐚 𝐚𝐝𝐚𝐥𝐚𝐡 { 𝟔𝟎° , 𝟏𝟐𝟎° }

𝟑) 𝐓𝐞𝐧𝐭𝐮𝐤𝐚𝐧 𝐡𝐢𝐦𝐩𝐮𝐧𝐚𝐧 𝐩𝐞𝐧𝐲𝐞𝐥𝐞𝐬𝐚𝐢𝐚𝐧 𝐩𝐞𝐫𝐬𝐚𝐦𝐚𝐚𝐧 𝐜𝐨𝐬 𝟓𝐱 = 𝟎,𝟔𝟒𝟐𝟕 𝐮𝐧𝐭𝐮𝐤 𝟎° ≤ × ≤𝟑𝟔𝟎°.

𝐉𝐚𝐰𝐚𝐛

𝐂𝐨𝐬 𝟓𝐱 = 𝟎,𝟔𝟒𝟐𝟕

𝐂𝐨𝐬 𝟓𝐱 = 𝐜𝐨𝐬 𝟓𝟎° 

•𝟓𝐱 = ±𝟓𝟎° + 𝐤. 𝟑𝟔𝟎°

• 𝐱 = ±𝟏𝟎° ± 𝐤. 𝟕𝟐°

𝐔𝐧𝐭𝐮𝐤 𝐤=𝟎

𝐗𝟏=𝟏𝟎°+𝟎.𝟕𝟐°=𝟏𝟎°

𝐗𝟐=-𝟏𝟎°+𝟎.𝟕𝟐°=-𝟏𝟎°(𝐭𝐢𝐝𝐚𝐤 𝐦𝐞𝐦𝐞𝐧𝐮𝐡𝐢)

𝐔𝐧𝐭𝐮𝐤 𝐤=𝟏

𝐗𝟑=𝟏𝟎°+𝟏.𝟕𝟐°=𝟖𝟐°

𝐗𝟒=-𝟏𝟎°+𝟏.𝟕𝟐=𝟔𝟐°

𝐔𝐧𝐭𝐮𝐤 𝐤=𝟐

𝐗𝟓=𝟏𝟎°+𝟐.𝟕𝟐°=𝟏𝟓𝟒°

𝐗𝟔=-𝟏𝟎°+𝟐.𝟕𝟐°=𝟏𝟑𝟒°

𝐉𝐚𝐝𝐢, 𝐡𝐢𝐦𝐩𝐮𝐧𝐚𝐧 𝐩𝐞𝐧𝐲𝐞𝐥𝐞𝐬𝐚𝐢𝐚𝐧𝐧𝐲𝐚 𝐚𝐝𝐚𝐥𝐚𝐡 {𝟏𝟎°, 𝟔𝟐°, 𝟖𝟐° ,𝟏𝟑𝟒° ,𝟏𝟓𝟒°}.

𝟒) 𝐓𝐞𝐧𝐭𝐮𝐤𝐚𝐧 𝐇𝐢𝐦𝐩𝐮𝐧𝐚𝐧 𝐩𝐞𝐧𝐲𝐞𝐥𝐞𝐬𝐚𝐢𝐚𝐧 𝐭𝐚𝐧 𝐱+𝟏 = 𝟎 𝐮𝐧𝐭𝐮𝐤 𝟎 <_ 𝐱 <_𝟐π

𝐉𝐚𝐰𝐚𝐛𝐚𝐧:

𝐃𝐢𝐤𝐞𝐭𝐚𝐡𝐮𝐢 

π = 𝟏𝟖𝟎°

𝟐π = 𝟑𝟔𝟎°

𝐉𝐚𝐰𝐚𝐛

𝐓𝐚𝐧 𝐱+𝟏=𝟎 

𝐭𝐚𝐧 𝐱= -𝟏

𝐌𝐞𝐧𝐜𝐚𝐫𝐢 𝐭𝐚𝐧 𝐱=-𝟏

𝐭𝐚𝐧 𝟏 = 𝟒𝟓°

𝐭𝐚𝐧 -𝟏 = 𝟏𝟖𝟎°-𝟒𝟓°

           = 𝟏𝟑𝟓°

𝐓𝐚𝐧 𝐗 = 𝟏𝟑𝟓°

𝐗 = 𝟏𝟑𝟓°+𝐤.𝟏𝟖𝟎°

𝐊= 𝟎...𝐗= 𝟏𝟑𝟓°...𝐗 =𝟏𝟑𝟓°/𝟏𝟖𝟎°π

                                = 𝟑/𝟒π

𝐊= 𝟎...𝐗=𝟑𝟏𝟓°....𝐗 = 𝟑𝟏𝟓°/𝟏𝟖𝟎°π

                                =𝟕/𝟒π

𝐌𝐚𝐤𝐚 𝐇𝐏={𝟑/𝟒π,𝟕/𝟒π}

𝟓) Tentukan penyelesaian dari 

Cos x = Sin 70° untuk 0° ≤ x ≤ 360°

Jawab:

Diubah dari. *Cos x = Sin 70°* menjadi *Sin 70° = Cos 20°*

Cos x = Cos 20°

x = 20° + k . 360°

k = 0 —> x = 20°

k = 1 —> x = 20° + 360° = 380° (TM)

atau

x = –20° + k . 360°

k = 0 —> x = –20° (TM)

k = 1 —> x = –20° + 360° = 340°

Jadi, himpunan penyelesaiannya adalah { 20° , 340° }