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

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

Apr 25, 2015

#2
+33586
+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°

.

Apr 26, 2015

#1
+126679
+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°

Apr 25, 2015
#2
+33586
+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