Buktikan bahwa cosx1+sinx+1+sinxcosx=cosx2 dan cari semua besar sudut antara −180∘ dan 180∘ yang merupakan solusi dari persamaan (1+sinx)2+cos2x=4cosx(1+sinx).

Pertanyaan

Buktikan bahwa $\frac{1+\sin\;x}{\cos\;x}+\frac{\cos\;x}{1+\sin\;x}=\frac2{\cos\;x}$ dan cari semua besar sudut antara $-180^\circ$ dan $180^\circ$ yang merupakan solusi dari persamaan $\left(1+\sin x\right)^2+\cos^2x=4\cos\;x\;(1+\sin\;x)$ .

Pembahasan Soal:

$begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 1 plus sin space x over denominator cos space x end fraction plus fraction numerator cos space x over denominator 1 plus sin space x end fraction end cell equals cell fraction numerator 1 plus sin space x over denominator cos space x end fraction plus fraction numerator cos space x over denominator 1 plus sin space x end fraction cross times fraction numerator 1 minus sin space x over denominator 1 minus sin space x end fraction end cell row blank equals cell fraction numerator 1 plus sin space x over denominator cos space x end fraction plus fraction numerator cos space x space left parenthesis 1 minus sin space x right parenthesis over denominator 1 minus sin squared x end fraction end cell row blank equals cell fraction numerator 1 plus sin space x over denominator cos space x end fraction plus fraction numerator cos space x space left parenthesis 1 minus sin space x right parenthesis over denominator cos squared x end fraction end cell row blank equals cell fraction numerator 1 plus sin space x over denominator cos space x end fraction plus fraction numerator 1 minus sin space x over denominator cos space x end fraction end cell row blank equals cell fraction numerator 1 plus sin space x plus 1 minus sin space x over denominator cos space x end fraction end cell row blank equals cell fraction numerator 2 over denominator cos space x end fraction space left parenthesis terbukti right parenthesis end cell end table end style$

$begin mathsize 12px style open parentheses 1 plus sin x close parentheses squared plus cos squared x equals 4 cos space x space left parenthesis 1 plus sin space x right parenthesis end style$
$begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 1 plus sin x close parentheses squared plus cos squared x minus 4 space cos space x space left parenthesis 1 plus sin space x right parenthesis end cell equals 0 row cell open parentheses 1 plus sin x close parentheses squared plus cos squared x minus 4 space cos space x space left parenthesis 1 plus sin space x right parenthesis end cell equals 0 row cell 1 plus 2 space sin space x plus sin squared x plus cos squared x minus 4 space cos space x minus 4 space cos space x space sin space x end cell equals 0 row cell 1 plus 2 space sin space x plus 1 minus 4 space cos space x minus 4 space cos space x space sin space x end cell equals 0 row cell 2 space sin space x minus 4 space cos space x minus 4 space cos space x space sin space x plus 2 end cell equals 0 row cell sin space x minus 2 space cos space x minus 2 space cos space x space sin space x plus 1 end cell equals 0 row cell left parenthesis sin space x plus 1 right parenthesis left parenthesis negative 2 space cos space x plus 1 right parenthesis end cell equals 0 end table end style$

$begin mathsize 12px style sin space x equals negative 1 space atau space cos space x equals 1 half end style$

$sin space x equals negative 1$

$table attributes columnalign right center left columnspacing 0px end attributes row cell sin space x end cell equals cell negative 1 end cell row cell sin space x end cell equals cell sin space 270 degree end cell row x equals cell 270 plus k.360 degree end cell row blank blank blank row k equals 0 row blank rightwards double arrow cell x equals 270 degree plus 0.360 end cell row blank equals cell 270 degree space left parenthesis tidak space memenuhi right parenthesis end cell row blank blank blank row x equals cell left parenthesis 180 minus 270 right parenthesis plus k.360 end cell row blank equals cell negative 90 degree plus k.360 end cell row blank blank blank row k equals 0 row x equals cell negative 90 degree plus 0.360 degree end cell row blank equals cell negative 90 degree end cell row blank blank blank row k equals 1 row x equals cell negative 90 degree plus 1.360 degree end cell row blank equals cell 270 degree space left parenthesis tidak space memenuhi right parenthesis end cell end table$

$cos space x equals 1 half$

$table attributes columnalign right center left columnspacing 0px end attributes row cell cos space x end cell equals cell 1 half end cell row cell cos space x end cell equals cell cos space 60 degree end cell row x equals cell 60 degree plus k.360 degree end cell row blank blank blank row k equals 0 row blank rightwards double arrow cell x equals 60 degree plus 0.360 degree end cell row blank equals cell 60 degree end cell row blank blank blank row x equals cell negative 60 degree plus k.360 degree end cell row blank blank blank row k equals 0 row blank rightwards double arrow cell x equals negative 60 degree plus 0.360 degree end cell row blank equals cell negative 60 degree end cell end table$

Jadi, himpunan penyelesaian dari persamaan trigonometri tersebut adalah $open curly brackets negative 60 degree comma space minus 90 degree comma space 60 degree close curly brackets$ .

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

I. Sutiawan

Mahasiswa/Alumni Universitas Pasundan

Terakhir diupdate 07 Oktober 2021