#2**+5 **

You could also solve this algebraically as follows. Knowing that tan θ = sinθ/cosθ we can write:

cosθ - sinθ/cosθ = 0

Multiply through by cosθ

cos^{2}θ - sinθ = 0

Also, cos^{2}θ = 1 - sin^{2}θ so

1 - sin^{2}θ - sinθ = 0

Rearrange:

sin^{2}θ + sinθ - 1 = 0

This is a quadratic in sinθ with solutions -(1-√5)/2 and -(1+√5)/2. The second result is not a valid solution because it is outside the range ±1.

So taking the arcsine of the other result we get

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}^{\!\!\mathtt{-1}}{\left({\mathtt{\,-\,}}{\frac{\left({\mathtt{1}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{5}}}}\right)}{{\mathtt{2}}}}\right)} = {\mathtt{38.172\: \!707\: \!627\: \!012^{\circ}}}$$

There is another value at 180°-38.173° = 141.827°

.

Alan Apr 26, 2015

#1**+5 **

cosΘ - tanΘ =0 add tanΘ to both sides

cosΘ = tanΘ

The easiest way to solve this - IMHO - is with a graph.......here it is....

https://www.desmos.com/calculator/wvqydtk4mt

The points of intersection occur at about 38.2° and at about 141.8°

And converting to rads we have ....about .6667 rads and about 2.475 rads on (0, 2pi)

CPhill Apr 25, 2015

#2**+5 **

Best Answer

You could also solve this algebraically as follows. Knowing that tan θ = sinθ/cosθ we can write:

cosθ - sinθ/cosθ = 0

Multiply through by cosθ

cos^{2}θ - sinθ = 0

Also, cos^{2}θ = 1 - sin^{2}θ so

1 - sin^{2}θ - sinθ = 0

Rearrange:

sin^{2}θ + sinθ - 1 = 0

This is a quadratic in sinθ with solutions -(1-√5)/2 and -(1+√5)/2. The second result is not a valid solution because it is outside the range ±1.

So taking the arcsine of the other result we get

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}^{\!\!\mathtt{-1}}{\left({\mathtt{\,-\,}}{\frac{\left({\mathtt{1}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{5}}}}\right)}{{\mathtt{2}}}}\right)} = {\mathtt{38.172\: \!707\: \!627\: \!012^{\circ}}}$$

There is another value at 180°-38.173° = 141.827°

.

Alan Apr 26, 2015