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

2 Answers
Sep 26, 2016

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....

Sep 26, 2016

(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!