How to solve #cosx-2sinx=1# for #0<=x<=2pi#?

2 Answers
Feb 16, 2018

#color(blue)(0,pi,2pi,5.356,4.070)color(white)(88)# ( 3 .d.p.)

Explanation:

#cosx-2sinx=1#

Add #2sinx# to both sides:

#cosx=1+2sinx#

Square both sides:

#cos^2x=(1+2sinx)^2#

Subract #(1+2sinx)^2# from both sides:

#cos^2x-(1+2sinx)^2=0#

Using identity:

#color(red)(bbsin^2x+cos^2x=1)#

#color(red)bb(cos^2x=1-sin^2x)#

#1-sin^2x-(1+2sinx)^2=0#

Expand bracket:

#1-sin^2x-(1+4sinx+4sin^2x)=0#

#1-sin^2x-1-4sinx-4sin^2x=0#

#-5sin^2x-4sinx=0#

#5sin^2x+4sinx=0#

Let #u=sinx#

Then:

#5u^2+4u=0#

Factor:

#u(5u+4)=0=>u=0 and u=-4/5#

But #u=sinx#

#sinx=0#

#sinx=-4/5#

#x=arcsin(sinx)=arcsin(0)=>x=0,pi,2pi#

#x=arcsin(sinx)=arcsin(-4/5)=>x=arcsin(-4/5)+2pi#

#x= pi-arcsin(-4/5)#

#color(blue)(0,pi,2pi,5.356,4.070)color(white)(88)# ( 3 .d.p.)

Feb 16, 2018

#x = 0^@#, and #x = 233.1^@#

Explanation:

cos x - 2sin x = 1
Call #tan a = sin a/(cos a) = 2# --> #a = 63^@43#
After cross multiplication -->
#cos x.cos a - sin a.sin x = cos a = cos 63^@43 = 0.45#
Using sum identity, we get:
#cos (x + a) = 0.45#
There are 2 solutions:
#x + a = +- 63^@43#
a. #x + 63.43 = 63^@43# --> #x = 0^@#
b. #x + 63.43 = - 63^@43# --> #x = - 126.87#, or
#x = 360 - 126.9 = 233^@13#
Check.
x = 0 --> cos x = 1 --> -2sin x = 0 -->
cos x - 2sin x = 1 . proved
x = 233.13 --> cos x = - 0.60 --> 2sin x = (2)(-0.8) = - 1.60 -->
--> cos x - 2sin x = - 0.6 + 1.6 = 1. Proved