Expert: Abe Mantell - 8/19/2010

Question
Can you please show me how to solve the following problems. 1.) cotxcos^2x=2cotx   2.)cosx+sinxtanx=2  3.)2sin^2x+3cosx-3=0. Thank you!

Answer
Hello Miranda,

1. cot(x)cos^2(x)=2cot(x)
==> cot(x)cos^2(x)-2cot(x)=0
==> cot(x)[cos^2(x)-2]=0
==> cot(x)=0 OR cos^2(x)=2
==> cot(x)=0 at x=pi/2, -pi/2, etc...
.   i.e. pi/2 + or - integer multiples of pi
.   cos^2(x)=2 has no real solutions (since cos^2(x)<=1)

2. cos(x)+sin(x)tan(x)=2, write tan(x) as sin(x)/cos(x)
==> cos(x)+sin(x)sin(x)/cos(x)=2, now multiply by cos(x)
==> cos^2(x)+sin^2(x)=2cos(x), but cos^2(x)+sin^2(x)=1
==> 1=2cos(x) ==> cos(x)=1/2
==> x=pi/6 or -pi/6...or any integer multiple of 2pi added or
.   subtracted...

3. 2sin^2(x)+3cos(x)-3=0, write sin^2(x) as 1-cos^2(x)
==> 2[1-cos^2(x)]+3cos(x)-3=0
==> 2-2cos^2(x)+3cos(x)-3=0
==> 2cos^2(x)-3cos(x)+1=0, which factors
==> [2cos(x)-1][cos(x)-1]=0
==> 2cos(x)=1 OR cos(x)=1
==> cos(x)=1/2 OR cos(x)=1
==> cos(x)=1/2 has the same solutions as in the previous
.   problem, and cos(x)=1 has x=0, 2pi, -2pi, 4pi, -4pi, ...

OK?

Abe