# C2 trig changing the intervals

For this question:
Find the solutions of the equation Sin(3x-15)=0.5 for the interval 0<x<180

When I want to change the intervals, do I minus 15 and then multiply by 3 or multiply by 3 and then minus 15?
If you want to find all those solutions, why don't you just do inverse sin and then solve for 'x' and make sure you give the values within your interval?
Original post by claret_n_blue
If you want to find all those solutions, why don't you just do inverse sin and then solve for 'x' and make sure you give the values within your interval?

Because you can save solving for x by changing the interval in the first place
Original post by scientific222
For this question:
Find the solutions of the equation Sin(3x-15)=0.5 for the interval 0<x<180
When I want to change the intervals, do I minus 15 and then multiply by 3 or multiply by 3 and then minus 15?

Just so that there is an answer here, you would multiply by 3 and then minus 15. Think of it as changing the interval, which in this case is 0<x<180, and making x match up with what's inside the trig function, Sin(3x-15). To do that you would multiply each term in the inequality by 3 and then minus 15, giving -15<3x-15<525. Hope this helps.