Question

If `3sec^2x - 4 = 0`, then what is the value of `x`?

Answer 1

Use the fact that `secx = 1/cosx`.

`3/cos^2x - 4 = 0`

`3/cos^2x = 4`

`3 = 4cos^2x`

`3/4 = cos^2x`

`cosx = +- sqrt(3)/2`

By the 30-60-90 special triangle, we have solutions of

`x = pi/6, (5pi)/6, (7pi)/6, (11pi)/6`

Now note that if you add `pin` with `n = 1` to the first two solutions you get the last two solutions.

The point of adding the `+pin` at the end is to account for the fact that trigonometric functions (e.g. sine, cosine, tangent, cosecant, secant and cotangent) all are periodic. That's to say, they repeat to infinity in both positive and negative directions.

Hence, if you picked any integral value of `n`, it should satisfy the equation, no matter how large or small.

Let `n = 9106`.

Then `x = pi/6 + 9106(pi) = (pi + 54636pi)/6 = (54637pi)/6`

If we check in the initial equation, we realize that this indeed is a solution.

Hopefully you understand now!

Answer 2

`3sec^2x-4=0`

`=>3(1+tan^2x)=4`

`=>3tan^2x=4-3=1`

`=>tanx=pm1/sqrt3`

When `tanx=1/sqrt3=tan(pi/6)`

then `x=npi+pi/6."where "n inZZ`

When `tanx =-1/sqrt3=-tan(pi/6)`

`=>tanx=tan(pi-pi/6)=tan((5pi)/6)`

`=>x=npi+(5pi)/6."where "n inZZ`

Otherwise

`3sec^2x-4=0`

`=>3sec^2x=4`

`=>secx=pm2/sqrt3`

`=>cosx=pmsqrt3/2`

When

`=>cosx=sqrt3/2=cos(pi/6)`

`=>x=2npipmpi/6" where "n in ZZ`

When

`=>cosx=-sqrt3/2=cos(pi-pi/6)`

`=>x=2npipm(5pi)/6" where "n in ZZ`