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

The general solution for 4cos(2t)+1=3,0, pi/2 is No Solution for t\in\mathbb{R}

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

4 cos (2 t) + 1 = 3, 0, \frac{π}{2}

No Solution for t \in R

Solution steps

4 cos (2 t) + 1 = 3, 0, \frac{π}{2}

Move

1

to the right side

4 cos (2 t) = 2

Divide both sides by

4

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

General solutions for

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

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

Solve

2 t = \frac{π}{3} + 2 πn : t = \frac{π}{6} + πn

Solve

2 t = \frac{5 π}{3} + 2 πn : t = \frac{5 π}{6} + πn

t = \frac{π}{6} + πn, t = \frac{5 π}{6} + πn

Solutions for the range

0

No Solution for t \in R

A, cos (\frac{π}{3} - A) = 2 sin (A)

- 3 + 3 = 26 cos (θ)

sin (\frac{π}{2} - x) = - 1, 0^{\circ} \leq x \leq 36 0^{\circ}

2 sin (5 x) = - 1

4 sin^{2} (x) + 8 cos (x) - 7 = 0

What is the general solution for 4cos(2t)+1=3,0, pi/2 ?
The general solution for 4cos(2t)+1=3,0, pi/2 is No Solution for t\in\mathbb{R}