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Pertanyaan

Jika negative text π end text less or equal than x less or equal than text π  end textdan space 4 sin squared open parentheses x close parentheses plus 4 cos open parentheses x close parentheses equals 1 commatext maka end text space x equals....

  1. 1 half text π end text space dan space minus 1 half straight pi space

  2. 1 fourth text π end text space dan space 3 over 4 straight pi space

  3. 2 over 3 text π end text space dan space minus 2 over 3 straight pi space

  4. 5 over 6 text π end text space dan space minus 1 over 6 straight pi space

  5. 2 over 3 text π end text space dan space minus 1 third straight pi space

Pembahasan Soal:

Diketahui space 4 sin squared open parentheses x close parentheses plus 4 cos open parentheses x close parentheses equals 1 dan identitas trigonometri sin squared open parentheses x close parentheses plus cos squared open parentheses x close parentheses equals 1 maka

table attributes columnalign right center left columnspacing 0px end attributes row cell space 4 sin squared open parentheses x close parentheses plus 4 cos open parentheses x close parentheses end cell equals 1 row cell 4 open parentheses 1 minus cos squared open parentheses x close parentheses close parentheses plus 4 cos open parentheses x close parentheses end cell equals 1 row cell 4 minus 4 cos squared open parentheses x close parentheses plus 4 cos open parentheses x close parentheses end cell equals 1 row cell negative 4 cos squared open parentheses x close parentheses plus 4 cos open parentheses x close parentheses plus 4 minus 1 end cell equals 0 row cell negative 4 cos squared open parentheses x close parentheses plus 4 cos open parentheses x close parentheses plus 3 end cell equals 0 row cell 4 cos squared open parentheses x close parentheses minus 4 cos open parentheses x close parentheses minus 3 end cell equals cell 0 space open parentheses text kedua ruas dikali end text space minus 1 close parentheses end cell row cell open parentheses 2 cos open parentheses x close parentheses plus 1 close parentheses open parentheses 2 cos open parentheses x close parentheses minus 3 close parentheses end cell equals 0 end table

2 cos open parentheses x close parentheses equals negative 1 space space space a t a u space space 2 cos open parentheses x close parentheses equals 3 cos open parentheses x close parentheses equals negative 1 half space space space space space space space space space space space space space cos open parentheses x close parentheses equals 3 over 2 open parentheses t i d a k space m e m e n u h i close parentheses x equals open curly brackets negative 1 third straight pi comma 2 over 3 straight pi close curly brackets space space space space space space space space space space space space space space space space space space space


Karena nilai cos open parentheses x close parentheses equals 3 over 2 space space space space space space space space space space space space space space space maka x untuk nilai tersebut tidak memenuhi karena cos open parentheses x close parentheses greater than 1 comma sehingga himpunan penyelesaian dari space 4 sin squared open parentheses x close parentheses plus 4 cos open parentheses x close parentheses equals 1adalah x equals open curly brackets negative 1 third straight pi comma 2 over 3 straight pi close curly brackets.


Oleh karena itu, jawaban yang benar adalah E.

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 03 Juni 2021

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