How do you solve #sinx + cosx = 1 # over the interval 0 to 2pi? Trigonometry Right Triangles Trigonometric Functions of Any Angle 1 Answer A. S. Adikesavan Mar 9, 2016 0, #pi/2# and #2pi#.. Explanation: With t = tan #(x/2)#, use sin x = #(2t)/(1+t^2) and cos x = (1-t^2)/(1+t^2)#. Then, t#(t-1)# = 0. t = tan #(x/2)# = 0 gives x = 0, 2#pi#. t = tan#(x/2)# = 1 gives x = #pi/2#.. 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 4591 views around the world You can reuse this answer Creative Commons License