# trig

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Find all solutions of cos 2x + √2/2 =0.

a) + or - 3/8pi +Kpi
b) 2/3pi + Kpi, 5/3pi + Kpi
c) + or - 2/5 +Kpi
d) 2/3 pi + 2Kpi, 5/3pi + 2Kpi

I really need help on this question :(

• trig -

cos 2x + √2/2 =0
cos 2x = -√2/2
I know cos π/4 = √2/2 or cos 45° = +√2/2
but 2x could be in II or III
2x = π-π/4 or 2x = π + π/4
2x = 3π/4 or 2x = 5π/4
x = 3π/8 or x = 5π/8 ---> 67.5° or 112.5°

but the period of cos 2x = π

general solution
x = 3π/8 + kπ, x = 5π/8 + kπ

a) fits the first of my solutions, but does not include the 2nd part of the solution
the others are not correct.

BTW, you can illustrate my 2nd solution, which they don't have, on a calculator

e.g. let x = 112.5° + 17(180)° = 3172.5
LS = cos 2(3172.5) + √2/2
= -.707106781 + .707106781
= 0
= RS

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