Question

How do you solve `20sinxcosxsecx - 3cos^2x= 3sin^2x`?

Answer 1

`x=sin^(-1)(0.15)=8.627^o`, nearly, is the principal value.

The general value `x = (n(180)+(-1)^n8.627)^o, n=0, +-1, +-2, +-3. ...`

Explanation:

Use `sin 2x=2 sin x and cos x and sin^2x+cos^2x=1`.

Here, the given equation simplifies to

`sin x =3/20=0.15`. So,

`x=sin^(-1)(0.15)=8.627^o`, nearly, is the principal value.

The general value `x = (n(180)+(-1)^n8.627)^o, n=0, +-1, +-2, +-3....`

Answer 2

`(10 xx 2sinxcosx)/cosx - 3cos^2x = 3sin^2x`

`(20sinxcosx)/cosx - 3cos^2x= 3sin^2x`

`20sinx - 3cos^2x = 3sin^2x`

`20sinx = 3sin^2x + 3cos^2x`

`20sinx = 3(sin^2x + cos^2x)`

`20sinx = 3`

`sinx = 3/20`

`x~= 8.6˚ and 171.4˚`

Hopefully this helps!