Expert: Sherman D. - 12/16/2008

Question
I am stuck on 3 Trig Equations that I have tried over and over again but i cant get the correct solution.   please help.
1)2cosx-sinx=-2  for 0< x < 2pi
2)cos^2x+3sinxcosx+1=0 for 0< x < 4pi
3)3sin2x-1=0 for 0< x < 2pi

Answer
1.)
2cos(x) - sin(x) = -2  
for 0 < x < 2pi

2cos(x) = -2 + sin(x)
cos(x) = -1 + (1/2)sin(x)

sorry i couldn't really show you how to work this.

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2.)
cos(x)^2 + 3sin(x)cos(x) + 1 = 0
for 0 < x < 4pi

cos(x)^2 + 3sin(x)cos(x) + (sin(x)^2 + cos(x)^2) = 0
cos(x)^2 + 3sin(x)cos(x) + sin(x)^2 + cos(x)^2 = 0
2cos(x)^2 + 3sin(x)cos(x) + sin(x)^2 = 0
(2cos(x) + sin(x))(cos(x) + sin(x)) = 0

2cos(x) + sin(x) = 0
2cos(x) = -sin(x)
2 = -sin(x)/cos(x)
2 = -tan(x)
tan(x) = -2
x = tan^-1(-2)
x = 296°33'54.18" or 116°33'54.18

or

cos(x) + sin(x) = 0
cos(x) = -sin(x)
1 = -sin(x)/cos(x)
-tan(x) = 1
tan(x) = -1
x = tan^-1(-1)
x = 315° or 135°

ANS : x = about 296°33'54.18", about 116°33'54.18, 315°, or 135°

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3.)
3sin(2x) - 1 = 0
for 0 < x < 2pi
for 0 < x < 2pi

3sin(2x) - 1 = 0
3sin(2x) = 1
sin(2x) = (1/3)
2x = 19°28'16.39" or 160°31'43.61"
x = 9°44'8.2" or 80°15'51.8"

x = about 9°44'8.2" or about 80°15'51.8"

for all answers to your problems, even ones that i may have left out, check www.quickmath.com, click on Solve, which is located under Equations, then just copy and past the problem, and click solve.