solve for theta: 2cos(theta) + 2sin(theta) = sqrt6 with theta between 0 and 2pi

Question

How do I solve this algebraically? Thanks!

Answer 1

2cosx + 2sinx = sqr6

divide by 2

cosx +sinx = sqr6/2,

replace cosx by sqr(1-sin^2x)

sqr(1-sin^2x) = (sqr6)/2 - sinx

square both sides

1-sin^2x = 6/4 -sqr6)(sinx) + sin^2(x)

2sin^2(x) -sqr6(sinx) +1/2 = 0

divide by 2

sin^2(x) -sqr6/2(sinx) + 1/4 = 0

use the quadratic formula

sinx = sqr6/4 + or - (1/2)sqr(6/4-1) = about .258 or .966

use a calculator with an inverse sine function

x = theta = about 15, 75, 105 or 165 degrees