# Сonvert trigonometric equation to quadratic equation

I am struggling to know how to transform this equation as it involves more than one type of trigonometric function, I know how to do it with one repeated function.

Question:

$$\sin^2 \theta/2$$ + $$\sin \theta$$ + $$1$$ = $$0$$

In must be transform in pure quadratic form:

$$t^2$$ + $$t$$ + $$1$$ = $$0$$

• Note that the trigonometric equation has at least the solution $\theta=-\pi/2$, while the trinomial $t^2+t+1$ has no real root (the discriminant is $<0$). – Jean-Claude Arbaut Apr 19 at 19:28
• Is it $\sin \frac{\theta}{2}$ or $\frac{\sin \theta}{2}$? – Vasya Apr 19 at 19:53

Observe that $$\sin \theta = 2 \sin \left (\frac {\theta} {2} \right ) \cos \left (\frac {\theta} {2}\right ).$$

• I think this doesn't help! – Dr. Sonnhard Graubner Apr 19 at 19:09
• Who says you? Put $t = \sin \left (\frac {\theta} {2} \right ).$ Then we have \begin{align*}t^2+2t \sqrt {1-t^2} + 1 - t^2 & = t^2 \implies (t+ \sqrt {1-t^2})^2 = t^2 \\ \implies t+ \sqrt {1-t^2} & = \pm t. \end{align*} – Dbchatto67 Apr 19 at 19:12
• It is not correct again, $$\cos(\frac{x}{2})=\pm\sqrt{1-\sin^2(\frac{x}{2})}$$ – Dr. Sonnhard Graubner Apr 19 at 19:14
• I don't understand what are you trying to say? – Dbchatto67 Apr 19 at 19:16
• Then read my comment. Do you know the $$\pm$$ sign? – Dr. Sonnhard Graubner Apr 19 at 19:17

This is what is sometime called a "trick question" since it does not require you to solve a quadratic equation.

Notice that if $$\sin^2\left(\dfrac{\theta}{2}\right)+\sin\theta+1=0$$, then it follows that $$\sin^2\left(\dfrac{\theta}{2}\right)=-\sin\theta-1\ge0$$. So it must be the case that $$\sin\theta\le-1$$. So $$\sin\theta=-1$$.

Therefore one would think that $$\theta=-\dfrac{\pi}{2}+2\pi n$$. But this will contain extraneous solutions. For example, $$-\dfrac{\pi}{2}$$ is not a solution since

$$\sin^2\left(-\frac{\pi}{4}\right)+\sin\left(-\frac{\pi}{2}\right)+1\ne0$$

This raises the question "Are there any values of $$n$$ which yield a solution?"

$$\begin{eqnarray} &&\sin^2\left(-\frac{\pi}{4}+\pi n\right)+\sin\left(-\frac{\pi}{2}+2\pi n\right)+1\\ &=&-\frac{\sqrt{2}}{2}\cos(\pi n)+0-\cos(2\pi n)+0\\ &=&-\frac{\sqrt{2}}{2}\cos(\pi n)-1 \end{eqnarray}$$ But, since $$\cos(\pi n)=\pm 1$$ there are no values of $$n$$ which will give $$0$$.

So the equation has no solutions.

ADDENDUM If it is the case that OP intended to write $$\dfrac{\sin^2\theta}{2}+\sin\theta+1=0$$, that also has no solutions since

$$\begin{eqnarray} &&\sin^2\theta+2\sin\theta+2\\ &=&(\sin\theta+1)^2+1\ge1 \end{eqnarray}$$

SECOND ADDENDUM Here is an even faster way to show that there are no solutions.

Rewrite the equation using the identity $$\sin^2\left(\dfrac{\theta}{2}\right)=\dfrac{1-\cos\theta}{2}$$ to obtain

$$2\sin\theta-\cos\theta+3=0$$ $$\sqrt{5}\left(\frac{2}{\sqrt{5}}\sin\theta-\frac{1}{\sqrt{5}}\cos\theta\right)+3=0$$

Letting $$\sin\psi=\dfrac{1}{\sqrt{5}}$$ this can be re-written

$$\sqrt{5}\sin(\theta-\psi)+3=0$$

But $$\sqrt{5}\sin(\theta-\psi)+3\ge3-\sqrt{5}>0$$. So there are no solutions.