Let S = [-,]2 : + = 2.

Question

Let $S = \left\{	heta\in[-\pi,\pi]\setminus\{\pm	frac\pi2\} : \sin	heta	an	heta+	an	heta = \sin2	hetaight\}$. If $T = \displaystyle\sum_{	heta\in S}\cos2	heta$, then $T+n(S)$ is equal to:

Accepted Answer

$$\sin\theta\tan\theta+\tan\theta = \sin2\theta \implies \tan\theta(\sin\theta+1) = 2\sin\theta\cos\theta$$

Multiplying both sides by $\cos\theta$ (valid since $\theta\ne\pm\tfrac\pi2$):

$$\sin\theta(\sin\theta+1) = 2\sin\theta\cos^2\theta$$

Case $\sin\theta = 0$: $\theta = -\pi, 0, \pi$ (all valid, $\cos\theta\ne0$ there).

Case $\sin\theta \ne 0$: $\sin\theta+1 = 2\cos^2\theta = 2-2\sin^2\theta \implies 2\sin^2\theta+\sin\theta-1=0 \implies (2\sin\theta-1)(\sin\theta+1)=0$

Since $\sin\theta\ne0$, and $\sin\theta=-1$ is excluded ($\theta=-\tfrac\pi2\notin$ domain), we get $\sin\theta=\tfrac12 \implies \theta = \dfrac\pi6, \dfrac{5\pi}6$.

So $S = \left\{-\pi,0,\pi,\dfrac\pi6,\dfrac{5\pi}6\right\}$, $n(S)=5$.

$$T = \cos(-2\pi)+\cos0+\cos2\pi+\cos\tfrac\pi3+\cos\tfrac{5\pi}3 = 1+1+1+\tfrac12+\tfrac12 = 4$$

$T+n(S) = 4+5 = 9$

Solutions · 1