# Solve equation $1+\sin^2\theta=3\sin\theta\cos\theta \text { (given } \tan71 ^{\circ}34^{\prime}=3)$

Solve equation $$1+\sin^2\theta=3\sin\theta\cos\theta \text { (given } \tan71 ^{\circ}34^{\prime}=3)$$

My attempt is as follows:-

$$1-2\sin\theta\cos\theta+\sin^2\theta-\sin\theta\cos\theta=0$$ $$\left(\sin\theta-\cos\theta\right)^2+\sin\theta\left(\sin\theta-\cos\theta\right)=0$$ $$\left(\sin\theta-\cos\theta\right)(2\sin\theta-\cos\theta)=0$$ $$\tan\theta=1 \text { or } \tan\theta=\frac{1}{2}$$ $$\theta=n\pi+\frac{\pi}{4}$$

For $$\tan\theta=\dfrac{1}{2} \text { how to make use of given condition } \tan71 ^{\circ}34^{\prime}=3$$

## 2 Answers

Use $$3=\frac{1+\frac12}{1-1\cdot\frac12}=\tan(45^\circ+\theta)$$.

• but how did you do that, guessing or something else – user3290550 Oct 26 '19 at 16:53
• @user3290550 I solved $3=\frac{k+\frac12}{1-\frac12 k}$ so I could use this. – J.G. Oct 26 '19 at 16:55

As $$1+2+3=1\cdot2\cdot3$$

$$\arctan1+\arctan2+\arctan 3=n\pi$$

As $$\arctan(u)<\dfrac\pi2$$

$$n=1$$

Now $$\arctan1=\dfrac\pi4$$

$$\arctan2+\arctan\dfrac12=\arctan2+$$arccot$$2=\dfrac\pi2$$