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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)
1−cos2(3π2−x)=−cosx+23sin(x−π)
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(3π2−x)=−sinxsin(x−π)=−sinx1−sin2x=−cosx−23sinx∣cosx∣=−cosx−23sinx(1)−cosx=−cosx−23sinx⇒sinx=0⇒x=nπ,n∈N(2)cosx=−cosx−23sinx⇒23sinx=−2cosx⇒tanx=−13=tan(−π6)⇒x=−π6+kπ,k∈Z
Commented by jagoll last updated on 17/Mar/20
thank you sir
thankyousir

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