How do you solve #cos^2x - sin^2x=sqrt3/2#? Trigonometry Trigonometric Identities and Equations Solving Trigonometric Equations 1 Answer Nghi N. · dani83 Aug 12, 2015 Solve #cos^2 x - sin^2 x = sqrt3/2# General solution: # x = npi +- pi/12, " " n in ZZ # Explanation: Use trig identity: #cos 2a = cos^2 a - sin^2 a# #cos^2 x - sin^2 x = cos 2x = sqrt3/2# #2x = +- pi/6# # 2x = 2npi +- pi/6, " " n in ZZ # General solution: # x = npi +- pi/12, " " n in ZZ # 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 4380 views around the world You can reuse this answer Creative Commons License