So I decided to give an another shot to calculus after I finished my CS degree for six years, and I'm using Paul Dawkins's Calculus 1 ebook. I'm fairly at the beginning but there're certain points I can't wrap my head around. On the 48th page for "Solving Trig Equations with Calculators, Part II" first example he solvesÉ
5*cos(2x-1)=-3
for which he gets 2.2143. Going by his analogy for the unit circle since cosine is the "horizontal" the -3/5 is either in the second or third quadrants. However, instead of having PI-2.2143 he goes and uses 2PI-2.2143 which to me is in the fourth quadrant where cosine is positive instead of -3/5.
5*cos(2x-1)=-3
cos(2x-1)=-3/5
2x-1=arcos(-3/5)=2.2143
I'm obviously missing something I just can't see what exactly... cosine -3/5 with 2.2143 should be in the third quadrant, and there's one more solution no matter how I look at that unit circle which is PI-2.2143, where did that 2PI-2.2143 come from?
It's true that the equation is the same as $$ \cos{(2x-1)} = -\frac{3}{5} $$ Let's set $\alpha = \arccos{-\frac{3}{5}} \approx 2.2143 ~\text{rad}$. Now we have $$ \cos{(2x-1)} = \cos{\alpha} $$ The correct way to solve this equation is to remember that $$ \cos{X} = \cos{Y} \qquad \Leftrightarrow \qquad X = \pm Y + n2\pi, \qquad n \in \mathbb{Z} $$ Therefore, $$ \cos{(2x-1)} = \cos{\alpha} \qquad \Leftrightarrow \qquad 2x-1 = \pm \alpha + n2\pi, \qquad n \in \mathbb{Z} $$ or $$ x = \frac{1}{2} \pm \frac{ \alpha}{2} + n\pi, \qquad n \in \mathbb{Z} $$ So the thing to keep in mind is that there are infinitely many solutions and they are found by this method.