# Thread: tanxsinx = 3sinx - secx

1. ## tanxsinx = 3sinx - secx

Solve for x:

equal to or less than 360
equal to or more than 0

tanxsinx = 3sinx - secx

A problem I have been struggling with for ages.

2. Originally Posted by differentiate
Solve for x:

equal to or less than 360
equal to or more than 0

tanxsinx = 3sinx - secx

A problem I have been struggling with for ages.
not very nice, I agree.

$\displaystyle \tan{x}\sin{x} = 3\sin{x} - \sec{x}$

$\displaystyle \frac{\sin{x}}{\cos{x}} \cdot \sin{x} = 3\sin{x} - \frac{1}{\cos{x}}$

$\displaystyle \frac{\sin^2{x}}{\cos{x}} = \frac{3\sin{x}\cos{x} - 1}{\cos{x}}$

$\displaystyle \sin^2{x} = 3\sin{x}\cos{x} - 1$

$\displaystyle \sin^2{x} = 3\sin{x}\cos{x} - (\sin^2{x} + \cos^2{x})$

$\displaystyle 2\sin^2{x} - 3\sin{x}\cos{x} + \cos^2{x} = 0$

$\displaystyle (2\sin{x} - \cos{x})(\sin{x} - \cos{x}) = 0$

$\displaystyle 2\sin{x} = \cos{x}$ or $\displaystyle \sin{x} = \cos{x}$

for $\displaystyle \sin{x} = \cos{x}$ ...

$\displaystyle x = 45^{\circ}$ , $\displaystyle x = 225^{\circ}$

for $\displaystyle 2\sin{x} = \cos{x}$

$\displaystyle 4\sin^2{x} = cos^2{x}$

$\displaystyle 4\sin^2{x} = 1 - \sin^2{x}$

$\displaystyle 5\sin^2{x} - 1 = 0$

$\displaystyle \sin^2{x} = \frac{1}{5}$

$\displaystyle \sin{x} = \pm \sqrt{\frac{1}{5}}$

the solutions to this part were found using a calculator and looking at a graph ...

$\displaystyle x = \arcsin\left(\sqrt{\frac{1}{5}}\right)$

and

$\displaystyle x = \arcsin\left(\sqrt{\frac{1}{5}}\right) + 180^{\circ}$