I tried solve command to solve the following system of equation:

3*sin(x) + 4*cos(y) = 6

4*sin(y) + 3*cos(x) = 1

Code:

`solve({3*sin(x)+4*cos(y)=6, 4*sin(y)+3*cos(x)=1},{x,y})`

Calculation is in radian, and found no solution.

When using TI Nspire Cas, I have got answer with the warning "More solutions may exist. Try specifying appropriate lower and upper bounds and/or a guess. "

Any setting required to let calculator to get the right solution?

fsolve for numerical approximate solutions, solve is for exact symbolic.

Also, it is much easier to put your equation in the command line, and then do COPY on your keyboard to get the text. Then you can just paste it in and anyone can paste into their system to test. Otherwise people have to type manually everything out.

In CAS:

fsolve([(3*sin(x)+4*cos(y)) = 6,(4*sin(y)+3*cos(x)) = 1],[x,y],[0,0])

Result:

[0.799756530541,−0.276157754943]

Please resize your images before posting them. It would really be appreciated.

Thank you for your help.

Apologize for posting too big picture. I have adjusted my post accordingly.

The "solve" command solves any system of equations (trigonometric, polynomial, exponential, square root, various). There is only one remark that you should remember and use. If the system of equations has many solutions, you must provide the approximate solution search area (lower bound, upper bound). I solved this system of trigonometric equations using this principle. The calculator has found many solutions for me, according to my wishes. Fantastic. See this in the attached screenshot from my HP PRIME. Good luck.

[

attachment=8263][

attachment=8264]

(03-27-2020 10:33 AM)Jan 11 Wrote: [ -> ]The "solve" command solves any system of equations (trigonometric, polynomial, exponential, square root, various). There is only one remark that you should remember and use. If the system of equations has many solutions, you must provide the approximate solution search area (lower bound, upper bound). I solved this system of trigonometric equations using this principle. The calculator has found many solutions for me, according to my wishes. Fantastic. See this in the attached screenshot from my HP PRIME. Good luck.

I'm using derivative (f'=0) to see the monotonic areas [a, b] where to search the roots with different sign at the ends of the areas. These two together ensure

only one root in the area.