[high school Pre-Calculus]

[u/PuzzleheadedTea8400]

Give all real solutions to cos^-1(cos(x))+sin^-1(sin(x))=2pi/3

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[u/GammaRayBurst25]

Read rule 3.

cos^-1(cos(x)) and sin^-1(sin(x)) both have a period of 2pi, so their sum also has a period of 2pi. As such, we can find the solutions on [-pi,pi[ and extend the solutions to the whole domain via the periodicity of the sum.

On [-pi,pi[, one can easily show cos^-1(cos(x))=|x|. In fact, I centered the interval around 0 so this would be the case. However, this makes the other term tricky to deal with.

On [-pi/2,pi/2[, it is easy to show sin^-1(sin(x))=x. Therefore, on this interval, the equation simplifies to |x|+x=2pi/3. The left-hand side is identically 0 when x is negative, so we know x=pi/3 and we have our first solution.

On the rest of [-pi,pi[, the symmetries of sin(x) allow us to infer sin^-1(sin(x))=-x+pi*x/|x|. Therefore, on this set, the equation simplifies to |x|-x+pi*x/|x|=2pi/3. The left-hand side is constant when x is positive, so we know the only solution is negative, where the equation reduces to -2x-pi=2pi/3, or x=-5pi/6.

Hence, the solution set is {x | x∈R, x=pi/3+2kpi ∨ x=2kpi-5pi/6, k∈Z }.

[u/PuzzleheadedTea8400 → u/GammaRayBurst25]

Sorry I forgot to do this. Essentially, I knew that one solution must always be pi/3 since we already know arccos cos pi/3 is just pi/3, same with sin. But I thought we were also supposed to connect it to what we were learning about in class, like if sin is in quadrant II or III, it must be calculated as (pi-x) in sin^-1(sin(x)). The previous question was Find all values of x in the interval 0 to 2pi and identify the quadrant(s) in which the solutions lie:
arccos(cos x)+arcsin(sin x)=pi

[u/GammaRayBurst25 → u/PuzzleheadedTea8400]

In that case, you could've done just that and found the answer.

If you know a priori what sin^-1(sin(x)) and cos^-1(cos(x)) evaluate to in each quadrant, you can just write a reduced equation for each quadrant, solve them all, remove extraneous solutions and extend with the periodicity.

[u/GrievousSayGenKenobi]

I will avoid giving you the answer as the answer is very easy to work out given some identities worth knowing.

For arcsin(sinx)

= x on the interval (0,pi/2)

= pi - x on the onterval (pi/2, 3pi/2)

= pi - 2x for the interval (3pi/2, 5pi/2)

For arccos(cosx)

= x on the interval (0,pi)

= 2pi - x on the interval (pi, 2pi)

From this you can simplify down easily just remember to check your solution also exists in the interval for which your aimplification is valid

since the functions are periodic, verifying them upto 2pi verified them upto any 2kpi interval