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solve the first equation in each pair of equations or y and /or z. Then use the same strategy to solve the second equation for x in the interval 0<= x<= 2pie.

y^2= 1/3, tan^2x = 1/3

yz= y, tan x sec x = tanx

y^2 = 1/3
y = ± 1/√3

so tan^2 x = 1/3
tanx = ± 1/√3
so x could be in all 4 quadrants
knowing the ratio of the 30-60-90 triangle is 1:√3:2
x must be 30º or pi/6 radians
so
x = pi/6 in I
x = pi - pi/6 = 5pi/6 in II
x = pi + pi/6 = 7pi/6 in III
x = 2pi - pi/6 = 11pi/6 in IV

for tan x sec x = tanx
divide by tanx
secx = 1
cosx = 1
Where is the cosine graph 1 ?
I will let you finish it.

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