Question

Solve the following equations.
$sinx-4cosx+tanx=4.$

Answer 1

$sinx-4cosx+tanx=4$

$tanx-4cosx=4-sinx$

squaring both sides.

$(tanx-4cosx{)}^2=(4-sinx{)}^2$

$ta{n}^{2}x+16co{s}^{2}x-8cosx-tanx=16+si{n}^{2}x-8sinx$

$ta{n}^{2}x=16(1-co{s}^{2}x)si{n}^{2}x-8sinx+8sinx$

[As $tan\left(x\right)=\frac{sinx}{cosx}]$

$ta{n}^{2}x=17si{n}^{2}x$

$17si{n}^{2}x-\frac{si{n}^{2}x}{co{s}^{2}x}=0$

$\frac{si{n}^{2}x}{co{s}^{2}x}[17co{s}^{2}x-1]=0$

$ta{n}^{2}x[17co{s}^{2}x-1]=0$

$ta{n}^{2}x=0$ OR $co{s}^{2}x=1/17$

$tanx=0cosx=\pm \frac{1}{\sqrt{17}}$

$x=x\pi ,nϵIORx=co{s}^{-1}(\pm \frac{1}{\sqrt{17}})$