Question Number 104576 by bemath last updated on 22/Jul/20

2sin 2x −4sin ^2 x = 7cos 2x  with (π/2)< x < π   find sin 2x

Commented bybemath last updated on 22/Jul/20

thank you both

Answered by bobhans last updated on 22/Jul/20

⇔4sin x cos x−4sin ^2 x = 7(cos ^2 x−sin ^2 x)  4sin x (cos x−sin x) = 7(cos x−sin x)(cos x+sin x)  ⇒(sin x−cos x){4sin x−7cos x−7sin x)=0   { ((sin x=cos x (rejected))),((7cos x = −3sin x )) :}  tan x = −(7/3) → { ((sin x = (7/(√(58))))),((cos x = −(3/(√(58))))) :}  sin 2x = 2×(7/(√(58))) ×−(3/(√(58))) = −((21)/(29)) ★

Answered by OlafThorendsen last updated on 22/Jul/20

2sin2x−4((1−cos2x)/2) = 7cos2x  2sin2x−2 = 5cos2x  ⇒ 4sin^2 2x−8sin2x+4 = 25cos^2 2x  4sin^2 2x−8sin2x+4 = 25−25sin^2 2x  29sin^2 x−8sin2x−21 = 0  (sin2x−1)(29sin2x+21) = 0  sin2x = 1 ⇔ x = (π/4)+kπ, k∈Z  impossible, (π/2)<x<π  sin2x = −((21)/(29))  is a solution.

Answered by Dwaipayan Shikari last updated on 22/Jul/20

2sin2x−4sin^2 x=7cos^2 x−7sin^2 x  −7cos^2 x+4sinxcosx+3sin^2 x=0  −7cos^2 x+7sinxcosx−3sinxcosx+3sin^2 x=0  −7cosx(cosx−sinx)−3sinx(cosx−sinx)=0  (cosx−sinx)(7cosx+3sinx)=0  cosx=sinx  tanx=1  ((2tanx)/(tan^2 x+1))=sin2x=1(But it is not the solution)  7cosx=−3sinx  tanx=((−7)/3)  ((2tanx)/(tan^2 x+1))=(((−14)/3)/(((49)/9)+1))=((−14.3)/(58))=−((21)/(29))

Answered by ajfour last updated on 22/Jul/20

let  cos 2x=t  ⇒ 4(1−t^2 )=(7t+2−2t)^2   4−4t^2 =25t^2 +20t+4  ⇒  29t^2 +20t=0  ⇒  t=−((20)/(29)) ,  sin 2x = −(√(1−t^2 ))     sin 2x = −((√(49×9))/(29)) = −((21)/(29)) .