Solve for θ. tan⁡(2θ)−tan⁡(2θ)tan2(θ)=2

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Solve for θ.  tan⁡(2θ)−tan⁡(2θ)tan2(θ)=2

grbavit · Accepted Answer

Firstly, notice that tan⁡(2θ) appears in both terms on the left so let&#039;s factor that out. tan⁡(2θ)[1−tan2(θ)]=2 Now, we need the arguments of our tantan functions to be the same if we have any hopes of simplifying the left-hand side. So we&#039;ll use the double angle formula to convert that tan⁡(2θ) into an expression in terms of tan⁡(θ). So remember that the double angle formula for tantan is tan⁡(2θ)=2tan⁡(θ)1−(tan2(θ)) Using that, we get →tan⁡(θ)=1 Here we look at the unit circle. Note that tan⁡(θ)=1 means the same thing as sin⁡(θ)=cos⁡(θ) which means the same thing as the x component = the y component on the unit circle. That only occurs at two angles between 0 and 2π radians:π4  radians    and  5π4 radians.  Remember though that the angle doesn&#039;t necessarily have to be between 0 and 2π radians so we need to add 2πk to both of those angles. Hence the full solution is π4+2πk or 5π4+2πk for every integer k, or more succinctly   π4+πk for every integer k

Solve for θ. tan(2θ)−tan(2θ)tan^2(θ)=2

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