Solutions of $\tan(2x+3) = -1/2$

trigonometry

this is a silly question but here I go.

I was solving a question that required evaluating the derivative of an equation, which would result in finding a local min and a local max. The two points are in Quadrant 2 and 3,respectively.

The original equation is $\frac{\cos(2x+3)}{2e^x}$.

It came down to $\tan(2x+3) = -1/2$. After this, $\arctan(-1/2) = -0.463647..$. This lies in Q4, but I was wondering if I could use its complimentary angle (i.e., $2π – 0.463657$). The answer would be same, since $\tan(2π – \theta) = – \tan(\theta)$.

However, both yields different $x$ values and thus different answers.

I would be appreciated if you could explain the problem in my reasoning to me.

Best Answer

  1. Here is the correct solution:

    $$\tan(2x+3) = -1/2\\ \iff 2x-3=n\pi+\arctan\left(-\frac12\right)\:\:\text{for some } n\in\mathbb Z\\ \iff x=\frac{n\pi-3-\arctan\frac12}2\:\:\text{for some } n\in\mathbb Z$$

    finding a local min and a local max. The two required points are in Quadrant 2 and 3, respectively.

    Since the original function is differentiable on $\mathbb R,$ the required points are necessarily some of the above stationary points. And since $\tan$ is monotonic around its roots, the above stationary points alternate between minimum and maximum points.

    Plugging in $n=-1$ and $n=0$ gives $x=-3.30$ and $x=-1.73,$ which are a quadrant-2 minimum and a quadrant-3 maximum, respectively.

    Hence, the complete solution set is $$\left\{\frac{n\pi-3-\arctan\frac12}2\,\Bigg\vert\,n\equiv-1,0\pmod4\right\}.$$

  2. It came down to $\tan(2x+3) = -1/2$. After this, $\arctan(-1/2) = -0.463647..$. This lies in Q4,

    Yes, by definition and convention, $\arctanα$ lies in the interval $\left(-\frac\pi2,\frac\pi2\right),$ thus $$\tan\alpha=p\kern.6em\not\kern-.6em\implies\alpha=\arctan p;$$ for example, try $\alpha=\pi$ which is in neither quadrant $4$ nor $1.$

    but I was wondering if I could use its complimentary angle (i.e., $2π - 0.463657$). The answer would be same, since $\tan(2π - \theta) = - \tan(\theta)$.

    The complementary angle of $θ$ is $\displaystyle\left(\frac\pi2-θ\right)$ rather than $\left(2\pi+θ\right).$

    And yes, $$\tan\alpha=p\kern.6em\not\kern-.6em\implies\alpha=2\pi+\arctan p,$$ as you've just illustrated.

    However, both yields different $x$ values and thus different answers.

    Yes, your two suggested answers correspond to $n=0$ and $n=2,$ which give the points $x=-1.73$ and $x=1.41,$ respectively. Only the former (out of infinitely many points) satisfies this exercise's requirement.

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