SOLUTION: Im stuck. I need the exact solution from [o.2pi) tan^2(x)= 1-sec(x)

Algebra ->  Trigonometry-basics -> SOLUTION: Im stuck. I need the exact solution from [o.2pi) tan^2(x)= 1-sec(x)      Log On


   



Question 1025159: Im stuck. I need the exact solution from [o.2pi)
tan^2(x)= 1-sec(x)

Found 2 solutions by richard1234, KMST:
Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
Add 1 to both sides and use the identity



This is a quadratic in terms of sec x, which you can solve for. Remember that sec x = 1/cos x, and can only take on values whose absolute value is at least 1.

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
Using the trigonometric identities that we should all remember,
tan%28x%29=sin%28x%29%2Fcos%28x%29 and sec%28x%29=1%2Fcos%28x%29 ,
you can re-write the equation as
sin%5E2%28x%29%2Fcos%5E2%28x%29=1-1%2Fcos%28x%29 .
Multiplying both sides of the equal sign times cos%5E2%28x%29 ,
the equation above turns into
sin%5E2%28x%29=cos%5E2%28x%29-cos%28x%29 ,
and since sin%5E2%28x%29=1-cos%5E2%28x%29 ,
you can substitute and re-write it as
1-cos%5E2%28x%29%7D=cos%5E2%28x%29-cos%28x%29 <--> 0=2cos%5E2%28x%29-cos%28x%29-1 .
The last equation is a quadratic equation.
If you use y=cos%28x%29 , you can write it as
2y%5E2-y-1=0
You can solve for y using factoring, or completing the square, or the quadratic formula to get the solutions
system%28y=1%2C%22or%22%2Cy=-1%2F2%29 .
In the interval %22%5B+0+%2C%222pi%22%29%22 ,
cos%28x%29=1 happens only for highlight%28x=0%29 ,
while cos%28x%29=-1%2F2 happens only for highlight%28system%28x=2pi%2F3%2C%22or%22%2Cx=4pi%2F3%29%29 .