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Last minute maths question? Trigonometry equations!?

+2 votes
Solve for x in the interval 0<x<2pi

a) sin2^x + 2sinxcosx = 0
asked Feb 21, 2013 in TRIGONOMETRY by angel12 Scholar

1 Answer

+1 vote

Sin^2(x) + 2sin(x)cos(x) = 0

Add and subtracted by cos^2(x)

then  Sin^2(x) + 2sin(x)cos(x)+cos^(2)x- cos^2(x)=0

By mathematical index property: (a+b)^(2)=a^(2)+b^(2)+2(a)(b)

(sinx+cosx)^(2)-cos^2(x)=0

Add cos^2(x) to each side.

(sinx+cosx)^(2)=cos^2(x)

Take Square root each side.

√(sinx+cosx)^(2)=√[cos^(2)x]

sin(x)=cos(x)-cos(x)

sin(x)=o

x=0

answered Feb 22, 2013 by ferguson Rookie
edited Feb 23, 2013 by moderator

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