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!

How do you solve 20sinxcosxsecx - 3cos^2x= 3sin^2x20sinxcosxsecx - 3cos^2x= 3sin^2x?