How do you solve #2cosx-sinx+2cosxsinx=1# in the interval #0

Question

How do you solve #2cosx-sinx+2cosxsinx=1# in the interval #0<=x<=2pi#?

Quinn Anderson · Accepted Answer

#x=pi/3# or #x={5pi}/3#

# 2 cos x - sin x + 2 cos x sin x = 1 #

# 2 cos x - sin x = 1 - 2 cos x sin x #

Squaring, possibly introducing extraneous roots,

# 4 cos^2 x + sin ^2 x - 4 cos x sin x = 1 + 4 cos^2 x sin ^2 x -4 cos x sin x #

The cross term cancels, that's fortunate. Let's turn everything to cosine and let #c=cos ^2 x.#

# 4 c + 1 - c = 1 + 4 c (1 - c ) #

# -c = -4c^2#

#c(4c - 1) = 0#

# cos^2 x = c = 0 or cos^2 x = c= 1/4#

#cos x = 0 or cos x = pm 1/2 #

In the requested range, #x=pi/2, x={3pi}/2# from the first, #x =pi/3, {2pi}/3, {4pi}/3, {5pi}/3# from the second.

We squared an equation so we need to check these.

# 2 cos (pi/2) - sin (pi/2) + 2 cos (pi/2) sin (pi/2) = -1 quad# NOPE

# 2 cos (pi/3) - sin (pi/3) + 2 cos (pi/3) sin (pi/3) ##= 2(1/2) - \sqrt{3}/2 + 2(1/2) \sqrt{3}/2 = 1 quad sqrt#

#{5pi}/3# works too

# 2 cos({2pi}/3) - sin({2pi}/3) + 2 cos({2pi}/3) sin({2pi}/3)#

#= 2(-1/2) - (\sqrt{3}/2) + 2 (-1/2)(sqrt{3}/2) = -1 - sqrt{3} quad #NOPE

Similar for #{4pi}/3#

Final answer:

#x=pi/3# or #x={5pi}/3#

How do you solve #2cosx-sinx+2cosxsinx=1# in the interval #0&lt;=x&lt;=2pi#?