How do you solve secxcscx - 2cscx = 0secxcscx−2cscx=0?

1 Answer
Antoine
Jul 28, 2015

Factorize the left hand side and equate the factors to zero.
Then, use the notion that : secx=1/cosx" "secx=1cosx and cscx=1/sinxcscx=1sinx

Result : color(blue)(x=+-pi/3+2pi"k , k"in ZZ )

Explanation:

Factorizing takes you from
secxcscx-2cscx=0
to
cscx(secx-2)=0

Next, equate them to zero
cscx=0=> 1/sinx=0

However, there is no real value of x for which 1/sinx=0

We move on to secx-2=0

=>secx=2

=>cosx=1/2=cos(pi/3)

=>x=pi/3

But pi/3 is not the only real solution so we need a general solution for all the solutions.

Which is : color(blue)(x=+-pi/3+2pi"k , k "in ZZ )

Reasons for this formula:
We include -pi/3 because cos(-pi/3)=cos(pi/3)

And we add 2pi because cosx is of period 2pi

The General solution for any "cosine" function is :

x=+-alpha+2pi"k , k" in ZZ

where alpha is the principal angle which just an acute angle

For example : cosx=1=cos(pi/2)

So pi/2 is the principal angle!

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