What is the general solution for 3cos(2t)+4=1,0<= t<= pi ?

The general solution for 3cos(2t)+4=1,0<= t<= pi is t= pi/2

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3cos(2t)+4=1,0<= t<= pi

3 cos (2 t) + 4 = 1, 0 \leq t \leq π

t = \frac{π}{2}

Degrees

t = 9 0^{\circ}

Solution steps

3 cos (2 t) + 4 = 1, 0 \leq t \leq π

Move

4

to the right side

3 cos (2 t) + 4 = 1

Subtract

4

from both sides

3 cos (2 t) + 4 - 4 = 1 - 4

Simplify

3 cos (2 t) = - 3

3 cos (2 t) = - 3

Divide both sides by

3

3 cos (2 t) = - 3

Divide both sides by

3

\frac{3 cos ( 2 t )}{3} = \frac{- 3}{3}

Simplify

cos (2 t) = - 1

cos (2 t) = - 1

General solutions for

cos (2 t) = - 1

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}

2 t = π + 2 πn

2 t = π + 2 πn

Solve

2 t = π + 2 πn : t = \frac{π}{2} + πn

2 t = π + 2 πn

Divide both sides by

2

2 t = π + 2 πn

Divide both sides by

2

\frac{2 t}{2} = \frac{π}{2} + \frac{2 πn}{2}

Simplify

t = \frac{π}{2} + πn

t = \frac{π}{2} + πn

t = \frac{π}{2} + πn

Solutions for the range

0 \leq t \leq π

t = \frac{π}{2}

cos (2 θ) - 5 = - 9 sin (θ)

cos (x) = \frac{13}{19}

0 = 3 cos (2 x)

sin^{2} (x) - 9 cos (x) + 9 = 0

sinh (x) = \frac{8}{15}

What is the general solution for 3cos(2t)+4=1,0<= t<= pi ?
The general solution for 3cos(2t)+4=1,0<= t<= pi is t= pi/2