What is the general solution for cos(t)=(sin(t)= 1/2),0<= t< pi/2 ?

The general solution for cos(t)=(sin(t)= 1/2),0<= t< pi/2 is t= pi/3

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cos(t)=(sin(t)= 1/2),0<= t< pi/2

cos (t) = (sin (t) = \frac{1}{2}), 0 \leq t < \frac{π}{2}

t = \frac{π}{3}

Degrees

t = 6 0^{\circ}

Solution steps

cos (t) = (sin (t) = \frac{1}{2}), 0 \leq t < \frac{π}{2}

General solutions for

cos (t) = \frac{1}{2}

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

t = \frac{π}{3} + 2 πn, t = \frac{5 π}{3} + 2 πn

t = \frac{π}{3} + 2 πn, t = \frac{5 π}{3} + 2 πn

Solutions for the range

0 \leq t < \frac{π}{2}

t = \frac{π}{3}

cos (x) = \frac{10}{15}

\frac{sin ( b )}{350} = \frac{sin ( 15 3 ^{\circ} )}{536}

- 1 - 2 cos^{2} (x) + 3 cos (x) = 0

2 cos (2 x) + cos (x) = 0

sin (x) = - 0.435

What is the general solution for cos(t)=(sin(t)= 1/2),0<= t< pi/2 ?
The general solution for cos(t)=(sin(t)= 1/2),0<= t< pi/2 is t= pi/3