Solving trig equation $10\sin^2x + 10\sin x\cos x - \cos^2x - 2 = 0, -360 < x < 360$

Question

I’m having a hard time solving this trigonometric equation:

$$10\sin^2x + 10\sin x\cos x - \cos^2x - 2 = 0, -360 < x < 360$$

I’ve tried factoring but it didn’t work. Any help will be appreciated. …

Answer 1

If $\cos{x}=0$ then $\sin^2x=\frac{1}{5},$ which gives $\sin^2x+\cos^2x=\frac{1}{5},$ which is a contradiction.

Thus, $\cos{x}\neq0.$

Now, let $\tan{x}=t$.

Thus, $$10t^2+10t-1-2(1+t^2)=0$$ or $$8t^2+10t-3=0$$ or $$8t^2+12t-2t-3=0$$ or $$(4t-1)(2t+3)=0.$$ Can you end it now?

Answer 2

Set $x=2 \tan^{-1}(t)$ using a tan half angle substitution.

Then $\cos(x)=\frac{1-t^2}{1+t^2}$ and $\sin(x) = \frac{2 t}{1+t^2}$

The expression is now

$$ \frac{40 t^2}{(1+t^2)^2} + \frac{20 t (1-t^2)}{(1+t^2)^2} - \frac{(1-t^2)^2}{(1+t^2)^2}-2 = 0$$

$$ - \frac{ (t^2+8t-1) (3 t^2-4 t-3) }{(1+t^2)^2} =0 $$

$$ \begin{cases} t^2+8 t-1 =0 & \Rightarrow & t=-4 \pm \sqrt{17} \\ 3 t^2-4 t-3 =0 & \Rightarrow & t = \frac{2}{3}\pm \frac{\sqrt{13}}{3} \end{cases}\\ \begin{cases} x = \tan^{-1}\left( \frac{1}{4} \right) & & x = \tan^{-1}\left( \frac{1}{4} \right)-\pi \\ x = \tan^{-1}\left( \frac{2}{3} \right) + \frac{\pi}{2} & & x = \tan^{-1}\left( \frac{2}{3} \right) - \frac{\pi}{2} \end{cases} $$

Answer 3

First, change $10\sin^2x + 10\sin x\cos x - \cos^2x - 2 = 0 $ to $11\sin^2x + 10\sin x\cos x =3 $.

Next, use the double angle formulas to make this $3 =11(1-\cos(2x))/2 + 5\sin(2x) =11/2-11\cos(2x)/2 + 5\sin(2x) $ or, setting $2x = y$, $5/2 =11\cos(y)/2 - 5\sin(y) $ or $5 =11\cos(y) - 10\sin(y) $.

The right side is of the form $a\cos(y)+b\sin(y) $. Dividing by $c =\sqrt{a^2+b^2} =\sqrt{221} $ gives $u\cos(y)+v\sin(y) $ where $u = 11/c, v=-10/c$.

Choose $z$ such that $\sin(z) = u,\cos(z) = v$ by $\tan(z) = u/v =-11/10$ gives $u\cos(y)+v\sin(y) =\sin(z)\cos(y)+\cos(z)\sin(y) =\sin(z+y) $ so $5 =c\sin(z+y) $ or $y =\arcsin(5/c)-z $.

Using $\arcsin(x) =\arctan(x/\sqrt{1-x^2}) $,

$\begin{array}\\ \arcsin(5/c) &=\arctan((5/c)/\sqrt{1-(5/c)^2})\\ &=\arctan(5/\sqrt{c^2-25})\\ &=\arctan(5/\sqrt{196})\\ &=\arctan(5/14)\\ \end{array} $

so

$\begin{array}\\ y &=\arctan(5/14)-\arctan(-11/10)\\ &=\arctan(5/14)+\arctan(11/10)\\ &=\arctan(\frac{5/14+11/10}{1-(5/14)(11/10)})\\ &=\arctan(\frac{12}{5})\\ \end{array} $

Since $x = y/2$, $\tan(y) =\tan(2x) =\frac{2\tan(x)}{1-\tan^2(x)} =\frac{2t}{1-t^2} $ where $t = \tan(x)$.

Therefore $2t=(12/5)(1-t^2)$ or $t^2+5t/6= 1$ or $(t+5/12)^2 =1+(5/12)^2 =(13/12)^2 $ or $t =5/12 \pm 13/12 =3/2, -2/3 $.

Therefore (whew! - there probably is a much easier way) $x = \arctan(3/2)$ or $\arctan(-2/3) $.