Question 921917
Solve for x: cos(4x)-(1/2)sin(4x)=1, For[0,90˚]
let u=4x
cosu-sinu/2=1
2cosu-sinu=2
2cosu-2=sinu
2(cosu-1)=sinu
square both sides
4(cos^2u-2cosu+1)=sin^2u
4cos^2u-8cosu+4=1-cos^2u
5cos^2u-8cosu+3=0
(cosu-1)(5cosu-3)=0
cosu=1
u=0=4x
x=0
or
cosu=3/5
u≈53.13˚=4x
x≈13.28˚