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.


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

In must be transform in pure quadratic form:

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

  • $\begingroup$ 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 $<0$). $\endgroup$ – Jean-Claude Arbaut Apr 19 at 19:28
  • $\begingroup$ Is it $\sin \frac{\theta}{2}$ or $\frac{\sin \theta}{2}$? $\endgroup$ – Vasya Apr 19 at 19:53

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

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


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