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

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Answer 1 · 2016-09-26T14:43:36

#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. ...#

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 · 2016-09-26T14:58:01

#(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!