sin 2 theta + sin theta - tan theta = 0 sin2θ+sinθ−tanθ=0
2 sin theta cos theta + sin theta - sin theta / cos theta = 0 2sinθcosθ+sinθ−sinθcosθ=0
We can assume cos theta ne 0cosθ≠0. Let's multiply to clear the fraction.
2 sin theta cos^2 theta + sin theta cos theta - sin theta = 0 2sinθcos2θ+sinθcosθ−sinθ=0
sin theta ( 2 cos ^2 theta + cos theta - 1 ) = 0sinθ(2cos2θ+cosθ−1)=0
sin theta = 0 sinθ=0 or 2 cos ^2 theta + cos theta - 1 = 02cos2θ+cosθ−1=0
theta = 0^circ or 180^circ or 360^circθ=0∘or180∘or360∘ from the first one.
( 2 cos theta - 1)(cos theta + 1) = 0 (2cosθ−1)(cosθ+1)=0
cos theta = 1/2 or cos theta = -1 cosθ=12orcosθ=−1
The first is trig's go-to triangle. theta =60^circ or 300^circθ=60∘or300∘
The second is theta=180^circθ=180∘ which we already had.
Check:
theta =0 quad sin 0 + sin 0 - tan 0 = 0 quad sqrt
theta=60^circ quad sin 120+ sin 60-tan60=\sqrt{3}/2+\sqrt{3}/2- \sqrt{3} = 0 quad sqrt
theta=180^circ quad sin 360 + sin 180 - tan 180= 0+0-0=0 quad sqrt
theta =300^circ quad sin 600 + sin 300 - tan 300 = sin(-120)+sin(-60)-tan(-60) = (-\sqrt{3}/2) + (-\sqrt{3}/2) - (-\sqrt{3}) = 0 quad sqrt
