+0

what values for Θ (0 < Θ < 2 pi) satisfy the equation? cos Θ- tan Θ = 0

0
1377
2

what values for Θ (0 < Θ < 2 pi) satisfy the equation? cos Θ- tan Θ = 0

Guest Apr 25, 2015

#2
+26328
+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

cos2θ - sinθ = 0

Also, cos2θ = 1 - sin2θ so

1 - sin2θ - sinθ = 0

Rearrange:

sin2θ + 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

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

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

.

Alan  Apr 26, 2015
Sort:

#1
+78620
+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°

CPhill  Apr 25, 2015
#2
+26328
+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

cos2θ - sinθ = 0

Also, cos2θ = 1 - sin2θ so

1 - sin2θ - sinθ = 0

Rearrange:

sin2θ + 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

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

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

.

Alan  Apr 26, 2015

28 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details