All Questions Topic List
Trigonometry Questions
Previous in All Question Next in All Question
Previous in Trigonometry Next in Trigonometry
Question Number 102783 by bramlex last updated on 11/Jul/20
23−1=6sin(2θ−60o)−2sin(2θ−30o)⇒θ=?
Answered by 1549442205 last updated on 11/Jul/20
⇔6sin2θ×12−6cos2θ×32−(2sin2θ×32−2cos2θ×12)=23−1⇔(3−3)sin2θ−(33−1)cos2θ=23−1(1)Putx=tanθ.Wehave(1)⇔(3−3)2x1+x2−(33−1)1−x21+x2=23−1(2)2(3−3)x−(33−1)(1−x2)=(23−1)(1+x2)(6−23)x−33+1+(33−1)x2−23+1−(23−1)x2=03x2+2(3−3)x−(53−2)=0tanθ=x1=1.361080303⇔θ≈53°41′41tanθ=x2=−2.825181918⇔θ≈−70°30′30
Answered by bemath last updated on 11/Jul/20
Commented by floor(10²Eta[1]) last updated on 11/Jul/20
whytheeq.hassolutionifthatinequalityhappens?
Terms of Service
Privacy Policy
Contact: info@tinkutara.com