I was trying to solve this purely trigonometric equation:


I performed my usual tactic,which is to rewrite everything in terms of sin and cos, and solve the resultant equation. (Works 100% of the time,70% of the time) and got:


Which I could not solve, although if anybody CAN solve it I would love to hear how. So I input the original equation into a computer algebra system and it told me that it is an equivalent problem to the solution of this equation:


Now this was solvable with the same approach that failed on the original equation. I just converted it into a quartic polynomial problem at the end, after rewriting it in terms of sin and cos.

But a number of questions arose, the answers to which would greatly help my ability to solve trig equations, but for the purpose of this site I think i'll split them up into multiple posts each with a very precise focus, so that they will be more useful to other users.

Question: How do I know if a given trigonometric equation can be simplified? (of course im primarily concerned about non obvious cases)

and by


I think the example I gave is a good one for what I mean by simplified. Two things happened:

(1) The number of terms in the equation that contained trigonometric functions was reduced. it went from 3 in the original to only 2 in the computer algebra system's output

(2) The order (natural power) of any trig terms in the equations was reduced.

How do I know if it is possible to accomplish one or both of these goals, given a purely trigonometric equation? (note: im not asking for a general procedure to do so, just a method of determining if this can be done given a purely trigonometric equation. I will ask other questions that arose from this problem later. )

  • 3
    $\begingroup$ Why do you stick to sin and cos rather than using the identity $\sec^2 x = \tan^2 x + 1$ from the outset? You immediately get a quadratic in $\tan x$. $\endgroup$ Apr 23, 2023 at 22:33
  • 3
    $\begingroup$ Generally speaking, you don't know until you try. $\endgroup$ Apr 23, 2023 at 22:33
  • 1
    $\begingroup$ Agreeing with what @AdamRubinson says, I think this is why math is "difficult". You just need to put in the practice. $\endgroup$
    – Scene
    Apr 23, 2023 at 22:39
  • $\begingroup$ It's impossible in the general case as that's equivalent to solving the halting problem. $\endgroup$
    – corvus_192
    Apr 24, 2023 at 12:59

1 Answer 1



Start from here,

$$\begin{align} 1&=\sin^4(x)+3\sin(x)\cos^3(x)\\ \\ 1-\sin^4(x)&=3\sin(x)\cos^3(x)\\ \\ (1+\sin^2(x))\cos^2(x)&=3\sin(x)\cos^3(x)\\ \\ &\Rightarrow \cos(x)=0\tag{1}\\ \\ &\text{or}\\ \\ 1+\sin^2(x)&=3\sin(x)\cos(x)\\ \\ \cos^2(x)+2\sin^2(x)&=3\sin(x)\cos(x)\\ \\ 1+2\tan^2(x)&=3\tan(x)\\ \\ (\tan(x)-1)(2\tan(x)-1)&=0\\ \\ &\Rightarrow \tan(x)=1 ~~\text{or} ~~\tan(x)=\frac{1}2\tag{2} \end{align}$$

From (1), (2) we get:

$$x=\frac{\pi}2+n\pi, ~\frac{\pi}4+n\pi,~ \arctan\left(\frac{1}2\right)+n\pi, n\in\mathbb{Z}$$


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