TRIG HELP!?

Update : Correcgtion

if you meant

sqrt(3) * sin(t) = cos(t)

divide both sides by sin(t)

sqrt(3) = cos(t)/sin(t)

so this is

cot(t) = sqrt(3)

cos( pi/6) = sqrt(3)/2

sin(pi/6) = 1/2

so

cotan(pi/6) = sqrt(3)

since cotangent has period of pi

t = pi/6 and 7pi/6

sqrt(3) * sin(t) = cos(t)

Divide it by 2.

sqrt 3/2 sin t =1/2 cos t

Now replace the values

cos t cos 60 - sin 60 sin t=0

cos (t+60)=0=cos 90

This gives t+60=90, t=30 degrees.

sint = cost

sint = sin (90° - t)

t = 90° - t

t + t = 90°

2t = 90°

t = 45°.

√3.sin(t) = cos(t)

sin(t)/cos(t) = 1/√3

tan(t) = 1/√3

t = (π/6) + kπ ← where k is an integer

First case: k = 0 → t = π/6

Second case: k = 1 → t = (π/6) + π = 7π/6

Third case: k = 2 → t = (π/6) + 2π = 13π/6 → over 2π

→ Solution = { π/6 ; 7π/6 }

Ans. sqr(3)sin(t)=cos(t)=>sin(t)/cos(t)=1/sqr(3)

=>tan(t)=1/sqr(3)=>t=pi/3, pi+pi/3=4pi/3.