How do you solve 2cosx plus 2sinx equals sqrt2 for solutions between 0 and 2pi?

Answer 1

Not easily and, given the limitations of the pathetic browser that we have at our disposal, even less easy. But here goes:You are given that 2cosx + 2sinx = sqrt(2)

Consider "collapsing" the left hand side into a single trig function:

r*sin(x+a) = sqrt(2) ............................................. (A)

that is r*cosxsina + r*sinxcosa = sqrt(2)

comparing coefficients,

rsina = 2 and rcosa = 2

then

sina = 2/r = cosa

and since sin^2 + cos^2 = 1, you have r = 2*sqrt(2)

also, rsina/rcosa = 2/2 = 1 = tana which implies that a = pi/4.

Therefore, equation (A) is 2*sqrt(2)*sin(x+pi/4) = sqrt(2)

so that sin(x+pi/4) = 1/2.......................................(B)

Since 0<= x <= 2*pi

pi/4 < x+pi/4 < 2*pi+pi/4

So the solutions to (B) in the relevant domain are x+pi/4 = 5*pi/6 or 13*pi/6

therefore x = 5*pi/6-pi/4 or 13*pi/6-pi/4

that is x = 7pi/12 and 23pi/12.