How do you solve #(cos x - 1)(cos x - 4) = 0# over the interval 0 to 2pi? Trigonometry Right Triangles Trigonometric Functions of Any Angle 1 Answer Konstantinos Michailidis Feb 5, 2016 Hence #abs(cosx)<=1# we have that #cosx-1=0=>cosx=1# so we have that #x=0# or #x=2pi# Answer link Related questions How do you find the trigonometric functions of any angle? What is the reference angle? How do you use the ordered pairs on a unit circle to evaluate a trigonometric function of any angle? What is the reference angle for #140^\circ#? How do you find the value of #cot 300^@#? What is the value of #sin -45^@#? How do you find the trigonometric functions of values that are greater than #360^@#? How do you use the reference angles to find #sin210cos330-tan 135#? How do you know if #sin 30 = sin 150#? How do you show that #(costheta)(sectheta) = 1# if #theta=pi/4#? See all questions in Trigonometric Functions of Any Angle Impact of this question 2484 views around the world You can reuse this answer Creative Commons License