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Trigonometry

I'm really stuck on this at the moment

Solve these equations for values of x from 0 to 2pi

a) cos x + cos5x = 0
b) cos4x - cosx = 0
c) sin3x - sinx = 0
d) sin2x + sin3x = 0

what I did in a) was to do 2cos(3x)cos(2x) = 0 but had no idea how to work out what the many values of x were in pi,7

also with part b) I wasn't sure how to form the equation is it supposed to be 2sin(5x/2)sin(3x/2) and again like with the other 3 parts I didn;t know how to find the values of pi
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TeeEm
19
Original post by bl64
I'm really stuck on this at the moment

Solve these equations for values of x from 0 to 2pi

a) cos x + cos5x = 0
b) cos4x - cosx = 0
c) sin3x - sinx = 0
d) sin2x + sin3x = 0

what I did in a) was to do 2cos(3x)cos(2x) = 0 but had no idea how to work out what the many values of x were in pi,7

also with part b) I wasn't sure how to form the equation is it supposed to be 2sin(5x/2)sin(3x/2) and again like with the other 3 parts I didn;t know how to find the values of pi


the standard way of solving these is to use the identities for

sin(A) +/- sin(B)
cos(A) +/- cos(B)

which is given in all examining boards

PS for (c) and (d) there is a cleverer quicker way
Original post by bl64
I'm really stuck on this at the moment

Solve these equations for values of x from 0 to 2pi

a) cos x + cos5x = 0
b) cos4x - cosx = 0
c) sin3x - sinx = 0
d) sin2x + sin3x = 0

what I did in a) was to do 2cos(3x)cos(2x) = 0 but had no idea how to work out what the many values of x were in pi,7

also with part b) I wasn't sure how to form the equation is it supposed to be 2sin(5x/2)sin(3x/2) and again like with the other 3 parts I didn;t know how to find the values of pi


following from what you had we have either cos3x=0 or cos2x=0
cos⁡3x=0  ⟹  3x=π2,3π2,5π2.\cos3x=0 \implies 3x=\frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{2}. etc
so just take all of these values within the specified range.
Then do the same with cos2x=0

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