solve for x: cosecX+secX=2Square root 2
Once you get to
cos(y) = cos(2y)
2cos^2(y)-cos(y)-1 = 0
(2cosy+1)(cosy-1) = 0
cosy = -1/2 or 1
y = 2π/3 or 5π/3 or 0
x = y + π/4
...
Nice thought! Thank you!
yes, except my error:
y = 2π/3 or 4π/3 or 0
a shorter way.
1/sin(x)+1/cos(x)=2√2
multiply by sin(x) on both sides,
1+tan(x)=2√2 sin(x)
square both sides
(1+tan(x))²=8sin²(x)=8tan²(x)/sec²(x)
using 1+tan²(x)=sec²(x),
(1+tan²(x))(1+tan(x))=8tan²(x)
set t=tan(x), and expand
t^4+2t³-6t²+2t+1=0
Factor:
(1+t)²(t²+4t+1)=0
=>
t={-√3-2,√3-2,1(multiplicity 2)}
x=atan(t)={11π/12,-5π/12,π/4}
or, equivalently,
x=atan(t)={11π/12,19π/12,π/4}
That's why I love trig - there are so many ways to mix and match those functions, sometimes in totally unexpected ways!
Thank you rabin for entertaining the math tutors :)
Perhaps there is an easier way to solve this problem, but this is all I have for now.
Where there is no confusion, sinX=sin(x), cosX=cos(x)
We'll work on the left side first
cscX+secX
=1/sinX+1/cosX
=(cosX+sinX)/(sinX*cosX)
=2(cosX+sinX)/sin(2x) [use sin(2x)=2sin(x)cos(x)]
=2√2 (cos(x)sin(π/4)+sin(x)cos(π/4))/sin(2x) [use √2(cos(π/4)=1]
=2√2 (sin(x+π/4)/sin(2x) [ use sin(a+b)=sin(a)cos(b)+cos(a)sin(b) ]
=2√2 (sin(y+π/2)/sin(2y+π/2) [ substitute y=x-π/4 ]
=2√2 cos(y)/cos(2y) [use sin(x+π/2=cos(x)]
Since the right-hand side equals 2√2
we equate LHS=RHS
2√2 cos(y)/cos(2y)=2√2
=>
cos(y)=cos(2y).................(1)
Since (1) implies that y and 2y are on either side of an axis of symmetry, which coincides with 3y/2, so
3y/2=π, or 3y/2=-π
=>
y1=2π/3, or y2=-2π/3
=>
Recall that y=x-π/4 => x=y+π/4,
x1=y1+π/4= 2π/3+π/4=11π/12, or
x2=y2+π/4=-2π/3+π/4=-5π/12.