If cosx cos(pi/7) + sinx sin(pi/7) = -(sqrt(2))/2, then x can (check all that apply) (will give medal & fan! ) a. pi/4 + pi/7 + 2npi b. 5pi/4 + pi/7 + 2npi c. 7pi/4 + pi/7 + 2npi d. 3pi/4 + pi/7 + 2npi

c is the correct answer

step-by-step explanation:

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.