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Question Number 84868 by jagoll last updated on 17/Mar/20
1−cos2(3π2−x)=−cosx+23sin(x−π)
Answered by john santu last updated on 17/Mar/20
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
thankyousir
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