Question

Solve the following equations.
$6si{n}^{2}x+sinxcosx-co{s}^{2}x=2.$

Answer 1

$6si{n}^{2}x+sinxcosx-co{s}^{2}x=2$

$6si{n}^{2}x+sinxcosx-(1-si{n}^{2}x)=2$

$7si{n}^{2}x-3=-sinxcosx$

squaring both sides

$49si{n}^{4}x-42si{n}^{2}x+9=si{n}^{2}x-si{n}^{4}x$

$si{n}^{4}x-\frac{43}{50}si{n}^{2}x+\frac{9}{50}=0$

take $t=si{n}^{2}x$

$t^2-\frac{43}{50}t+\frac{9}{50}=0$

$t=\frac{43/50\pm \sqrt{(43/50{)}^2-\frac{9\times 4\times 1}{50}}}{2}$

$t=0.5$ OR $0.36$

$sinx=\pm \frac{1}{\sqrt{2}}$ OR

$sinx=\pm 0.06$

$\pi /4,3\pi /4,5\pi /4,7\pi /4x={3.44}^{\circ }+n\pi /2n\epsilon I$

by putting in $e{q}^n$

So,

$x=3\pi /4,5\pi /4$