How to solve 1+2cos^2(x+pi/6)=3sin(pi/3-x)?

2 Answers
Jun 12, 2018

x = pi/6 + 2kpi
x = (5pi)/6 + 2kpi
x = (3pi)/2 + 2kpi

Explanation:

1 + cos^2 (x + pi/6) = 3sin (pi/3 - x) (1)
Note.
3sin (pi/3 - x) = 3cos (pi/2 - (pi/3 - x) = 3cos (x + pi/6).
Call (x + pi/6) = t . The equation (1) becomes:
2cos^2 t - 3cos t + 1 = 0.
Solve this quadratic equation for cos t.
Since a + b + c = 0, use shortcut. The 2 real roots are:
cos t = 1 and cos t = c/a = 1/2

a. cos t = cos (x + pi/6) = 1
x + pi/6 = 2pi
x = 2pi - pi/6 = (5pi)/6 + 2kpi
b. cos (x + pi/6) = 1/2
x + pi/6 = +- pi/3
1. x + pi/6 = pi/3 --> x = pi/3 - pi/6 = pi/6 + 2kpi
2. x + pi/6 = (5pi)/3 (co-terminal to x = - pi/3)
x = (5pi)/3 - pi/6 = (9pi)/6 = (3pi)/2 + 2kpi

Jun 12, 2018

x={2kpi+-pi/6,kinZZ}uu{2kpi-pi/2,kinZZ}

Explanation:

We know that,

color(violet)((1)sintheta=cos(pi/2-theta)

color(blue)((2)costheta=1=cos0=>theta=2kpi+-0=2kpi, kinZZ

color(red)((3)costheta=cosalpha=>theta=2kpi+-alpha,kinZZ

Here,

1+2cos^2(x+pi/6)=3color(violet)(sin(pi/3-x)...toApply(1)

=>1+2cos^2(x+pi/6)=3color(violet)(cos[pi/2-(pi/3-x)]

=>1+2cos^2(x+pi/6)=3cos[pi/2-pi/3+x]

=>1+2cos^2(x+pi/6)=3cos(pi/6+x)

=>2cos^2(x+pi/6)-3cos(x+pi/6)+1=0

Let , cos(x+pi/6)=m

:.2m^2-3m+1=0

=>2m^2-2m-m+1=0

=>2m(m-1)-1(m-1)=0

=>(m-1)(2m-1)=0

=>m-1=0 or 2m-1=0

=>m=1 or m=1/2,where, m=cos(x+pi/6)

=>cos(x+pi/6)=1 or cos(x+pi/6)=1/2

(i)cos(x+pi/6)=1

=>color(blue)(x+pi/6=2kpi, kinZZ...toApply(2)

=>x=2kpi-pi/6, kinZZ

(ii)cos(x+pi/6)=1/2=cos(pi/3)

=>color(red)(x+pi/6=2kpi+-pi/3.kinZZ...toApply(3)

=>x+pi/6=2kpi+pi/3 or x+pi/6=2kpi-pi/3 ,kinZZ

=>x=2kpi+pi/3-pi/6 orx=2kpi-pi/3-pi/6 ,kinZZ

=>x=2kpi+pi/6 ,or x=2kpi-pi/2 ,kinZZ

Hence,

x=2kpi-pi/6 orx=2kpi+pi/6 orx=2kpi-pi/2 ,kinZZ

i.e. x={2kpi+-pi/6,kinZZ}uu{2kpi-pi/2,kinZZ}