+33658 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
sin360∘−1(−(1−√5)2)=38.172707627012∘
There is another value at 180°-38.173° = 141.827°
.
+130477 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)
+33658 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
sin360∘−1(−(1−√5)2)=38.172707627012∘
There is another value at 180°-38.173° = 141.827°
.
