How do you solve the equation #2cos^2x- 7cosx + 3 = 0#? Precalculus Polynomial Functions of Higher Degree Zeros 1 Answer Noah G Aug 30, 2016 #x = pi/3 + 2pin and x = (5pi)/3 + 2pin# Explanation: Solve by factoring. #2cos^2x - 6cosx - cosx + 3 =0# #2cosx(cosx - 3) - 1(cosx - 3) = 0# #(2cosx -1)(cosx - 3) = 0# #cosx= 1/2 and cosx = 3# #x = pi/3 + 2pin, x = (5pi)/3 + 2pin# Hopefully this helps! Answer link Related questions What is a zero of a function? How do I find the real zeros of a function? How do I find the real zeros of a function on a calculator? What do the zeros of a function represent? What are the zeros of #f(x) = 5x^7 − x + 216#? What are the zeros of #f(x)= −4x^5 + 3#? How many times does #f(x)= 6x^11 - 3x^5 + 2# intersect the x-axis? What are the real zeros of #f(x) = 3x^6 + 1#? How do you find the roots for #4x^4-26x^3+50x^2-52x+84=0#? What are the intercepts for the graphs of the equation #y=(x^2-49)/(7x^4)#? See all questions in Zeros Impact of this question 13042 views around the world You can reuse this answer Creative Commons License