How do you solve $(cos2x)(cos(pi/3)) + (sin2x)(sin(pi/3)) = (sqrt(3))/2$ , where domain is [0,2pi]?

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Answer 1 · 2015-05-31T01:32:17

Use the trig identity: cos (a - b) = cos a.cos b + sin a.sin b

$cos (2x - Pi/3) = (sqrt3)/2 = cos +- (pi/6)$

a. $2x - pi/3 = pi/6 -> 2x = (3pi)/6 = pi/2 -> x = pi/4$

b. $2x - pi/3 = (11pi)/6 -> 2x = (13pi)/6 -> x = (13pi)/12$

How do you solve (cos2x)(cos(pi/3)) + (sin2x)(sin(pi/3)) = (sqrt(3))/2(cos2x)(cos(pi/3)) + (sin2x)(sin(pi/3)) = (sqrt(3))/2, where domain is [0,2pi]?