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Tentukan nilai x yang memenuhi persamaan berikut. b. tan−1 x=cos−1 x

Pertanyaan

Tentukan nilai x yang memenuhi persamaan berikut.

b. tan to the power of negative 1 end exponent space x equals cos to the power of negative 1 end exponent space x 

A. Salim

Master Teacher

Mahasiswa/Alumni Universitas Pelita Harapan

Jawaban terverifikasi

Jawaban

 nilai x yang memenuhi persamaan tersebut adalah x equals fraction numerator square root of 2 square root of 5 minus 2 end root over denominator 2 end fraction.

Pembahasan

Ingat bahwa:

  •  tan open parentheses tan to the power of negative 1 end exponent space x close parentheses equals x  

Mencari nilai x:

table attributes columnalign right center left columnspacing 0px end attributes row cell tan to the power of negative 1 end exponent space x end cell equals cell cos to the power of negative 1 end exponent space x end cell row cell tan open parentheses tan to the power of negative 1 end exponent space x close parentheses end cell equals cell tan space open parentheses cos to the power of negative 1 end exponent space x close parentheses end cell row x equals cell tan space open parentheses cos to the power of negative 1 end exponent space x close parentheses end cell end table 

Misalkan m equals cos to the power of negative 1 end exponent space x sehingga 

table attributes columnalign right center left columnspacing 0px end attributes row m equals cell cos to the power of negative 1 end exponent space x end cell row cell cos space m end cell equals cell cos open parentheses cos to the power of negative 1 end exponent space x close parentheses end cell row cell cos space m end cell equals x row x equals cell cos space m end cell end table 

Maka  x equals tan space m,

Oleh karena itu

x equals x tan space m equals cos space m fraction numerator sin space m over denominator cos space m end fraction equals cos space m sin space m equals cos squared space m sin space m equals 1 minus sin squared space m sin squared space m plus sin space m minus 1 equals 0  

Dengan rumus kuadratik, didapat akar-akarnya adalah:

table attributes columnalign right center left columnspacing 0px end attributes row cell sin space m subscript 1 comma thin space 2 end subscript end cell equals cell fraction numerator negative b plus-or-minus square root of b squared minus 4 a c end root over denominator 2 a end fraction end cell row cell sin space m subscript 1 comma thin space 2 end subscript end cell equals cell fraction numerator negative 1 plus-or-minus square root of 1 squared minus 4 times thin space 1 times open parentheses negative 1 close parentheses end root over denominator 2 times thin space 1 end fraction end cell row blank equals cell fraction numerator negative 1 plus-or-minus square root of 1 plus 4 end root over denominator 2 times thin space 1 end fraction end cell row blank equals cell fraction numerator negative 1 plus-or-minus square root of 5 over denominator 2 end fraction end cell row cell sin space m subscript 1 end cell equals cell fraction numerator negative 1 plus square root of 5 over denominator 2 end fraction equals 0 comma 61803 horizontal ellipsis space comma end cell row cell sin space m subscript 2 end cell equals cell fraction numerator negative 1 minus square root of 5 over denominator 2 end fraction equals negative 1 comma 61803 horizontal ellipsis end cell end table  

Mencari cos space m dengan pythagoras,

Untuk Error converting from MathML to accessible text. 

table attributes columnalign right center left columnspacing 0px end attributes row cell sin space m subscript 1 end cell equals cell fraction numerator negative 1 plus square root of 5 over denominator 2 end fraction comma space end cell row blank blank blank row cell sisi space depan end cell equals cell negative 1 plus square root of 5 space end cell row cell sisi space miring end cell equals 2 row cell sisi space samping end cell equals cell square root of open parentheses 2 close parentheses squared minus open parentheses negative 1 plus square root of 5 close parentheses squared end root end cell row blank equals cell square root of 4 minus open parentheses square root of 5 minus 1 close parentheses squared end root end cell row blank equals cell square root of 4 minus open parentheses 1 minus 2 square root of 5 plus 5 close parentheses end root end cell row blank equals cell square root of 4 minus open parentheses 6 minus 2 square root of 5 close parentheses end root end cell row blank equals cell square root of 2 square root of 5 minus 2 end root end cell row blank blank blank row cell cos space m end cell equals cell samping over miring end cell row blank equals cell fraction numerator square root of 2 square root of 5 minus 2 end root over denominator 2 end fraction end cell end table   

Untuk table attributes columnalign right center left columnspacing 0px end attributes row blank blank sin end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank m end table subscript 2 equals table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator negative 1 minus square root of 5 over denominator 2 end fraction end cell end table tidak memenuhi. 

Didapat table attributes columnalign right center left columnspacing 0px end attributes row cell cos space m end cell equals cell fraction numerator square root of 2 square root of 5 minus 2 end root over denominator 2 end fraction end cell end table.

Karena x equals cos space m, sehingga

x equals fraction numerator square root of 2 square root of 5 minus 2 end root over denominator 2 end fraction 

Jadi, nilai x yang memenuhi persamaan tersebut adalah x equals fraction numerator square root of 2 square root of 5 minus 2 end root over denominator 2 end fraction.

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