Friday
March 6, 2015

# Homework Help: Math - Solving Trig Equations

Posted by Anonymous on Saturday, November 24, 2007 at 7:07pm.

What am I doing wrong?

Equation:
sin2x = 2cos2x

90 and 270

....

My Work:

2sin(x)cos(x) = 2cos(2x)
sin(x) cos(x) = cos(2x)
sin(x) cos(x) = 2cos^2(x) - 1
cos(x) (+/-)\sqrt{1 - cos^2(x)} = 2cos^2(x) - 1
cos^2(x)(1 - cos^2(x)) = 4cos^4(x) - 4cos^2(x) + 1
5cos^4(x) - 5cos^2(x) + 1 = 0
cos(x) = (+/-) 0.85065080835215
cos(x) = (+/-)0.52573111211905
• Math - Solving Trig Equations - bobpursley, Saturday, November 24, 2007 at 7:33pm

wouldn't it be easier to solve for 2x then divide it by two?

sin2x=2cos2x

sin2x/cos2x=2
tan 2x=2

2x=63.43 deg
x= .... you do it.

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