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Jika x 1 ​ dan x 2 ​ memenuhi 2 sin x + sec x - 2 tan x - 1 = 0, maka nilai sin + cos yang mungkin adalah ....

Jika dan memenuhi 2 sin x + sec x - 2 tan x - 1 = 0, maka nilai sin begin mathsize 14px style straight x subscript 1 end style + cos begin mathsize 14px style straight x subscript 2 end style yang mungkin adalah ....

  1. begin mathsize 12px style 4 over 5 end style

  2. begin mathsize 12px style 3 over 4 end style

  3. begin mathsize 12px style 4 over 3 end style

  4. begin mathsize 12px style 3 over 2 end style

  5. 2

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A. Rizky

Master Teacher

Mahasiswa/Alumni Universitas Negeri Malang

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begin mathsize 14px style 2 space sin space invisible function application straight x plus sec invisible function application space straight x minus 2 space tan space invisible function application straight x minus 1 equals 0  2 space sin space invisible function application straight x plus fraction numerator 1 over denominator cos space invisible function application straight x end fraction minus fraction numerator 2 space sin invisible function application space straight x over denominator cos space invisible function application straight x end fraction minus 1 equals 0  2 space sin space invisible function application straight x minus 1 minus open parentheses fraction numerator 2 space sin invisible function application space straight x minus 1 over denominator cos space invisible function application straight x end fraction close parentheses equals 0  open parentheses 2 space sin invisible function application space straight x minus 1 close parentheses open parentheses 1 minus fraction numerator 1 over denominator cos space invisible function application straight x end fraction close parentheses equals 0    Maka colon  2 space sin space invisible function application straight x equals 1 rightwards double arrow sin invisible function application space straight x equals 1 half  1 minus fraction numerator 1 over denominator cos space invisible function application straight x end fraction equals 0 rightwards double arrow fraction numerator 1 over denominator cos space invisible function application straight x end fraction equals 1 rightwards double arrow cos invisible function application space straight x equals 1  Sehingga  sin invisible function application space straight x subscript 1 plus cos space invisible function application straight x subscript 2 equals 1 half plus 1 equals 3 over 2 end style

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Himpunan penyelesaian dari sin 2 x + 3 sin x + 2 = 0 adalah ...

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