Question

Solve the following equations.
$tan3x+cos6x=1.$

Answer 1

$tan\left(3x\right)+cos6x=1$

We know that,

$cos2\theta =\frac{1-ta{n}^2\theta }{1+ta{n}^2\theta }$

Here, $\theta =3x$

$\therefore cos6x=\frac{1-ta{n}^{2}3x}{1+ta{n}^{2}3x}$

Now, $tan3x+\frac{1+-ta{n}^{2}3x}{1+ta{n}^{2}3x}=1$

let $tan3x=t$

$y+\frac{1-y^2}{1+y^2}=1$

$\Rightarrow y^3+y+1-y^2=1+y^2$

$\Rightarrow y^3-2y^2+y=0$

$\Rightarrow y(y-1{)}^2=0$

$\Rightarrow y=1,0$

$tan\theta =tan3x=1,0(\alpha ,\frac{\pi }{4})$

General solvtion $n\pi +\alpha =3x$

$x=\frac{n\pi }{3}+\frac{\alpha }{4\left(3\right)},\frac{n\pi }{3}$

soluvtions are $\frac{5\pi }{12},\frac{9\pi }{12},\frac{13\pi }{12}...$

$0,\frac{\pi }{3},\frac{2\pi }{3},\pi ,\frac{4\pi }{3}...$