How do you solve $sin(x+60°) + cos(x+30°) = 1/2$ in the range 0° < x < 360°?

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Answer 1 · 2015-05-02T03:35:55

$x = 73.22^@$ and $x = 360 - 73.22 = 286.78^@$

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

$sin (x + 60) = (1/2).sin x + ((sqrt3)/2).cos x$

$cos (x + 30) = cos x.((sqrt3)/2) - (1/2).sin x$

Adding the 2 expressions, we get:

$(sqrt3).cos x= 1/2 -> cos x = 1/(2sqrt3) = (sqrt3)/6$

-> $x = 73.22^@$ and $x = 360 - 73.22 = 286.78^@$

How do you solve sin(x+60°) + cos(x+30°) = 1/2sin(x+60°) + cos(x+30°) = 1/2 in the range 0° < x < 360°?