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Question Number 84868 by jagoll last updated on 17/Mar/20

(√(1−cos^2 (((3π)/2)−x))) = −cos x+2(√3) sin (x−π)

1cos2(3π2x)=cosx+23sin(xπ)

Answered by john santu last updated on 17/Mar/20

cos (((3π)/2)−x) = −sin x  sin (x−π) = −sin x  (√(1−sin^2 x)) = −cos x−2(√3) sin x  ∣cos x∣ = −cos x−2(√3) sin x  (1) −cos x = −cos x −2(√3)sin x  ⇒ sin x = 0 ⇒ x = nπ , n ∈N  (2) cos x = −cos x−2(√3) sin x  ⇒2(√3) sin x = −2cos x   ⇒ tan x = −(1/((√3) )) = tan (−(π/6))  ⇒x = −(π/6)+ kπ , k∈Z

cos(3π2x)=sinxsin(xπ)=sinx1sin2x=cosx23sinxcosx=cosx23sinx(1)cosx=cosx23sinxsinx=0x=nπ,nN(2)cosx=cosx23sinx23sinx=2cosxtanx=13=tan(π6)x=π6+kπ,kZ

Commented by jagoll last updated on 17/Mar/20

thank you sir

thankyousir

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