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}

Solution steps

3 cos (2 t) + 4 = 1, 0 \leq t \leq π

Move

4

to the right side

3 cos (2 t) = - 3

Divide both sides by

3

cos (2 t) = - 1

General solutions for

cos (2 t) = - 1

2 t = π + 2 πn

Solve

2 t = 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