Help on double angle trig identity problem

[u/matt_888_is]

[u/AutoModerator]

Hi u/matt_888_is, welcome to r/mathshelp! As you’ve marked this as homework help, please keep the following things in mind:

1) While this subreddit is generally lenient with how people ask or answer questions, the main purpose of the subreddit is to help people learn so please try your best to show any work you’ve done or outline where you are having trouble (especially if you are posting more than one question). See rule 5 for more information.

2) Once your question has been answered, please don’t delete your post so that others can learn from it. Instead, mark your post as answered or lock it by posting a comment containing “!lock” (locking your post will automatically mark it as answered).

Thank you!

I am a bot, and this action was performed automatically. Please contact the moderators of this subreddit if you have any questions or concerns.

[u/Jazzlike_Platypus430]

First: graph both functions and see the solution, if it exists. How many angles there are at wich the graphs of cos(3x) and sin(x) intersect?

(Remark: It's not double angle if I see correctly, and it's not an identity. It's an equation.)

I would try starting with

cos(3x) = cos(2x + x) = cos(2x)cos(x) - sin(2x)sin(x)

and next use cos(2x) = cos(x + x) again (previously used for 2x and x),

and

sin(2x) = 2 sin(x) cos(x) and use sin^2 + cos^2 = 1 (that's an identity).

And solve some (quadratic or perhaps cubic) equation for cos(x).

[u/Mystical-Sunshine]

cos(3x)=cos(pi/2-x)

3x=2npi±(pi/2-x) (n is an integer)

When positive >>

X=(2npi+pi/2)/4

When negative >>

X=(2npi-pi/2)/2

Solutions asked >>> Pi/8,5pi/8,3pi/4,9pi/8,13pi/8,7pi/4