SOLUTION: How do you solve 7=5+sec(4x+7pi/4) 5+csc(2x+pi/3)=(15+2(sqroot3)/3) Thanks a lot!

Algebra ->  Trigonometry-basics -> SOLUTION: How do you solve 7=5+sec(4x+7pi/4) 5+csc(2x+pi/3)=(15+2(sqroot3)/3) Thanks a lot!      Log On


   



Question 735714: How do you solve
7=5+sec(4x+7pi/4)
5+csc(2x+pi/3)=(15+2(sqroot3)/3)
Thanks a lot!

Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
How do you solve
7=5+sec(4x+7pi/4)
5+csc(2x+pi/3)=(15+2(sqrt(3)/3)
***
7=5+sec(4x+7pi/4)
7-5=sec(4x+7pi/4)
2=1/cos(4x+7pi/4)
cos(4x+7pi/4)=1/2
%284x%2B7pi%2F4%29=inverse+cos+%281%2F2%29=%28pi%2F3%29
4x=pi%2F3-7pi%2F4
LCD:12
48x=4pi-21pi
x=%28-17pi%2F48%29=-1.1126+%28radians%29
...
5+csc(2x+pi/3)=(15+2(sqrt(3)/3)
csc(2x+pi/3)=15+2(sqrt(3)/3)-5
csc(2x+pi/3)=10+2(sqrt(3)/3)≈11.1547
sin(2x+pi/3)=1/11.1547≈0.0896
2x+pi/3=inverse sin(0.0896)≈0.0897
2x=0.0897-pi/3≈-0.9575
x≈-0.4787 radians