How do you find the x intercepts of #y=sec^4((pix)/8)-4#? Trigonometry Graphing Trigonometric Functions General Sinusoidal Graphs 1 Answer salamat Jan 31, 2017 #x=4n-2#, #n=1,2,3,#... Explanation: #y=sec^4((pix)/8)-4# at #x# intercepts, #y=0# #sec^4((pix)/8)-4=0# #(sec^2((pix)/8)-2)(sec^2((pi x)/8)+2)=0# #(sec^2((pi x)/8)+2)=0#, #sec^2((pi x)/8)=-2#, cannot solve. #sec^2((pix)/8)-2=0#, #sec((pix)/8)=+-sqrt2# #(pix)/8=pi/4, (3pi)/4, (5pi)/4,...# #x=2,6,10,...# #x=4n-2#, #n=1,2,3,#... Answer link Related questions What does sinusoidal mean? Given any sinusoidal equation, how do you identify the type of transformations that are made? How do you graph any sinusoidal graph? What does the coefficients A, B, C, and D to the graph #y=D \pm A \cos(B(x \pm C))#? What is the period, amplitude, and frequency for the graph #f(x) = 1 + 2 \sin(2(x + \pi))#? What is the period, amplitude, and frequency for #f(x)=3+3 cos (\frac{1}{2}(x-frac{\pi}{2}))#? How do you graph #y=2+3 \sin(2(x-1))#? How do you graph #y=2 cos(-x)+3#? How do you graph #y=3cos(4x)#? How do you graph #y=(cos2x)/2#? See all questions in General Sinusoidal Graphs Impact of this question 2823 views around the world You can reuse this answer Creative Commons License