How do you solve #sin(x+60°) + cos(x+30°) = 1/2# in the range 0° < x < 360°? Trigonometry Trigonometric Identities and Equations Solving Trigonometric Equations 1 Answer Nghi N. · dani83 May 2, 2015 # x = 73.22^@# and #x = 360 - 73.22 = 286.78^@# Explanation: 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^@# Answer link Related questions How do you find all solutions trigonometric equations? How do you express trigonometric expressions in simplest form? How do you solve trigonometric equations by factoring? How do you solve trigonometric equations by the quadratic formula? How do you use the fundamental identities to solve trigonometric equations? What are other methods for solving equations that can be adapted to solving trigonometric equations? How do you solve #\sin^2 x - 2 \sin x - 3 = 0# over the interval #[0,2pi]#? How do you find all the solutions for #2 \sin^2 \frac{x}{4}-3 \cos \frac{x}{4} = 0# over the... How do you solve #\cos^2 x = \frac{1}{16} # over the interval #[0,2pi]#? How do you solve for x in #3sin2x=cos2x# for the interval #0 ≤ x < 2π# See all questions in Solving Trigonometric Equations Impact of this question 22109 views around the world You can reuse this answer Creative Commons License