<s>[INST] Find the general solution of $\theta$ for which the following quadratic equation is the square of a linear function. Find the general solution of $\theta$ for which the quadratic equation
$$\left(\sin\theta\right)x^2+(2\cos\theta)x+\dfrac{\cos\theta+\sin\theta}{2}$$ is the square of a linear function.
$$D=0$$
$$4\cos^2\theta-2\sin\theta\left(\sin\theta+\cos\theta\right)=0$$
$$2\cos^2\theta-\sin^2\theta-\sin\theta\cos\theta=0$$
$$2\cos^2\theta-2\sin\theta\cos\theta+\sin\theta\cos\theta-\sin^2\theta=0$$
$$2\cos\theta(\cos\theta-\sin\theta)+\sin\theta(\cos\theta-\sin\theta)=0$$
$$(\cos\theta-\sin\theta)(2\cos\theta+\sin\theta)=0$$
$$\tan\theta=1 \text { or } \tan\theta=-2$$
$$\theta=n\pi+\dfrac{\pi}{4} \text {  or  } \theta=\tan(2\pi-\tan^{-1}(2))$$
$$\theta=n\pi+\dfrac{\pi}{4} \text {  or  } \theta=n\pi+2\pi-\tan^{-1}(2)$$
$$\theta=n\pi+\dfrac{\pi}{4} \text {  or  } \theta=\pi(n+2)-\tan^{-1}(2)$$
$$\sin\theta\ne 0$$
$$\theta\ne m\pi \text { where $m \in$ I }$$ 
$$\theta=n\pi+\dfrac{\pi}{4} \text { can't be integral multiple of $\pi$ as } \theta=\dfrac{\pi(4n+1)}{4}$$
$$\theta=n\pi+\dfrac{\pi}{4} \text { is the valid solution }$$
$$\theta=\pi(n+2)-\tan^{-1}(2) \text { cannot be the integral multiple of $\pi$ as $\tan^{-1}(2)$ is not the integral multiple of $\pi$ } $$
$$\theta=\pi(n+2)-\tan^{-1}(2) \text { is the valid solution }$$
Hence $\theta=\pi(n+2)-\tan^{-1}(2) \text { or } \theta=n\pi+\dfrac{\pi}{4}$
But actual answer is $\theta= 2n \pi+\dfrac{\pi}{4} \text{or }  \theta =(2n+1)\pi - \tan^{-1}(2) \text { where $n \in$ I}$
I tried to find out the mistake but didn't get any breakthrough. What am I missing.
 [/INST] Your problem is that the discriminant tells you when there is a double root, not when the polynomial is a square. When $\sin \theta$ is negative the polynomial factorises in the form $-(ax+b)^2$ and is the negative of a square.
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