So i have this trigonometric equations: $$2\cos^2(x)+4\sin(x)+\cos(2x)=0$$
I have rewritten the expression and came up with $$(2\sin(x)-3)(1+2\sin(x))=0$$
Then i split the equation in two and got
$$\sin(x)=3/2,\quad\text{and}\quad\sin(x)=-1/2$$
Since $\sin(x)=-1/2$ is a standard angle and with respect to the period I got $x=\pi - \sin^{-1}(11\pi/6)$
The problem is that I can not figure out what angle $\sin(x)=(3/2)$.
I have tried using the Pythagorean theorem but since $\sin$ is $\text{opp}/\text{hyp}$, it tells me that something is wrong since the hypotenuse can not be shorter than the sides.
Does anyone have an idea of what I should do?
$\endgroup$ 85 Answers
$\begingroup$The equation $\sin(x) = \frac{3}{2}$ has no solution because $-1 \leq \sin(x) \leq 1$ for all $x$.
So you can just solve $\sin(x) = -\frac{1}{2}$ and you're done.
Extending to complex plane, $$\sin^{-1} \frac{3}{2}=\frac{\pi}{2}-2i\ln \left( \frac{1+\sqrt{5}}{2} \right)$$
This can be easily verified by using
$$\sin (x+yi) = \sin x \cosh y+i\cos x \sinh y$$
$\endgroup$ 2 $\begingroup$Solve $$\sin(x)=\frac{3}{2}$$ for $x$.
Rewrite the sine function in terms of exponential function: $$\frac{3}{2}=\sin(x)=\frac{e^{ix}-e^{-ix}}{2i};$$ multiply both sides by $2i$: $$3i=e^{ix}-e^{-ix};$$ subtract $3i$ from both sides: $$e^{ix}-3i-e^{-ix}=0;$$ multiply both sides by $e^{ix}$: $$\left(e^{ix}\right)^2-3ie^{ix}-e^{-ix}e^{ix}=0;$$ rewrite $e^{-ix}e^{ix}=1$: $$\left(e^{ix}\right)^2-3ie^{ix}-1=0;$$ substitute $u=e^{ix}$: $$u^2-3iu-1=0;$$ multiply both sides by $4$: $$4u^2-12iu-4=0;$$ rewrite $4u^2=\left(2u\right)^2$: $$\left(2u\right)^2-12iu-4=0;$$ add $4+9i^2$ to both sides: $$\left(2u\right)^2-12iu+9i^2=4+9i^2;$$ rewrite $9i^2=\left(3i\right)^2$: $$\left(2u\right)^2-12iu+\left(3i\right)^2=4+9i^2;$$ rewrite $\left(2u\right)^2-12iu+\left(3i\right)^2=\left(2u-3i\right)^2$: $$\left(2u-3i\right)^2=4+9i^2;$$ rewrite $4+9i^2=4-9=-5$: $$\left(2u-3i\right)^2=-5;$$ take the square roots of both sides: $$2u-3i=\pm i\sqrt{5};$$ add $3i$ to both sides: $$2u=3i\pm i\sqrt{5};$$ factor $i$ out: $$2u=i\left(3\pm\sqrt{5}\right);$$ divide both sides by $2$: $$u=\frac{i}{2}\left(3\pm\sqrt{5}\right);$$ substitute back $u=e^{ix}$: $$e^{ix}=\frac{i}{2}\left(3\pm\sqrt{5}\right);$$ rewrite $\frac{i}{2}\left(3\pm\sqrt{5}\right)=e^{\log\left(3i/2\pm i\sqrt{5}/2\right)}$: $$e^{ix}=e^{\log\left(3i/2\pm i\sqrt{5}/2\right)};$$ since $1=e^{2i\pi c_1}$ for arbitrary integer $c_1$, multiply the RHS by $e^{2i\pi c_1}$: $$e^{ix}=e^{\log\left(3i/2\pm i\sqrt{5}/2\right)}e^{2i\pi c_1};$$ rewrite $e^{\log\left(3i/2\pm i\sqrt{5}/2\right)}e^{2i\pi c_1}=e^{\log\left(3i/2\pm i\sqrt{5}/2\right)+2i\pi c_1}$: $$e^{ix}=e^{\log\left(3i/2\pm i\sqrt{5}/2\right)+2i\pi c_1};$$ eliminate exponentials: $$ix=\log\left(\frac{3\pm\sqrt{5}}{2}i\right)+2i\pi c_1;$$ divide both sides by $i$: $$x=\frac{1}{i}\log\left(\frac{3\pm\sqrt{5}}{2}i\right)+2\pi c_1;$$ rewrite $1/i=-i$: $$x=-i\log\left(\frac{3\pm\sqrt{5}}{2}i\right)+2\pi c_1;$$ rewrite $\log\left(\frac{3\pm\sqrt{5}}{2}i\right)=\log\left(\frac{3\pm\sqrt{5}}{2}\right)+\frac{i\pi}{2}$: $$x=-i\left(\log\left(\frac{3\pm\sqrt{5}}{2}\right)+\frac{i\pi}{2}\right)+2\pi c_1;$$ expand: $$x=-i\log\left(\frac{3\pm\sqrt{5}}{2}\right)-\frac{i^2\pi}{2}+2\pi c_1;$$ rewrite $i^2=-1$: $$x=\frac{\pi}{2}-i\log\left(\frac{3\pm\sqrt{5}}{2}\right)+2\pi c_1.$$
From the whole set of solutions, the principal inverse sine value is delivered by setting $c_1=0$ and choosing the positive argument of $\log$: $$\arcsin\left(\frac{3}{2}\right)=\boxed{\frac{\pi}{2}-i\log\left(\frac{3+\sqrt{5}}{2}\right)},$$ where $\log$ is the natural logarithm and $i$ is the imaginary unit.
$\endgroup$ $\begingroup$Hint $-2 \leq 2 \sin(x) \leq 2$... So how can this be equal to 3?
$\endgroup$ $\begingroup$If you get $xy = 0$ then you can conclude one or the other is equal to 0. There is UTTERLY NO reason to assume they BOTH have to be 0.
Consider $x = 27; y = 0$ then $xy = 0$ so $x = 0$ or $y = 0$ but $x \ne 0$. That's NOT a contradiction or a problem at all as only ONE of the values has to be zero. They don't both have to 0.
So you got $(2\sin x - 3)(1 + 2\sin x) = 0$. So ONE of those terms (not BOTH) is equal to zero. As $\sin x \le 1$ it's impossible for $\sin x = 3/2$ so $2\sin x - 3 \ne 0$.
That's NOT a problem. There is no reason on earth you should have assumed it did.
However as it doesn't equal 0, we know the other term MUST and we know $\sin x = -1/2$.
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