What is the general solution for 4sin(2t)=4cos(t) ?

The general solution for 4sin(2t)=4cos(t) is t= pi/2+2pin,t=(3pi)/2+2pin,t= pi/6+2pin,t=(5pi)/6+2pin

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Popular >

4sin(2t)=4cos(t)

Solution

4 sin (2 t) = 4 cos (t)

Solution

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

Degrees

t = 9 0^{\circ} + 36 0^{\circ} n, t = 27 0^{\circ} + 36 0^{\circ} n, t = 3 0^{\circ} + 36 0^{\circ} n, t = 15 0^{\circ} + 36 0^{\circ} n

Solution steps

4 sin (2 t) = 4 cos (t)

Subtract

4 cos (t)

from both sides

4 sin (2 t) - 4 cos (t) = 0

Rewrite using trig identities

- 4 cos (t) + 4 sin (2 t)

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

= - 4 cos (t) + 4 \cdot 2 sin (t) cos (t)

Simplify

= - 4 cos (t) + 8 sin (t) cos (t)

- 4 cos (t) + 8 cos (t) sin (t) = 0

Factor

- 4 cos (t) + 8 cos (t) sin (t) : 4 cos (t) (2 sin (t) - 1)

- 4 cos (t) + 8 cos (t) sin (t)

Rewrite

8

2 \cdot 4

= - 4 cos (t) + 2 \cdot 4 sin (t) cos (t)

Factor out common term

4 cos (t)

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

4 cos (t) (2 sin (t) - 1) = 0

Solving each part separately

cos (t) = 0 or 2 sin (t) - 1 = 0

cos (t) = 0 : t = \frac{π}{2} + 2 πn, t = \frac{3 π}{2} + 2 πn

cos (t) = 0

General solutions for

cos (t) = 0

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}{2} + 2 πn, t = \frac{3 π}{2} + 2 πn

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

2 sin (t) - 1 = 0 : t = \frac{π}{6} + 2 πn, t = \frac{5 π}{6} + 2 πn

2 sin (t) - 1 = 0

Move

1

to the right side

2 sin (t) - 1 = 0

Add

1

to both sides

2 sin (t) - 1 + 1 = 0 + 1

Simplify

2 sin (t) = 1

2 sin (t) = 1

Divide both sides by

2

2 sin (t) = 1

Divide both sides by

2

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

Simplify

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

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

General solutions for

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

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

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

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

Combine all the solutions

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

Popular Examples

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1+2sin^2(t)=-5cos(t)

1 + 2 sin^{2} (t) = - 5 cos (t)

2sin(2x)-3-6/(sin(2x)-1)=0

2 sin (2 x) - 3 - \frac{6}{sin ( 2 x ) - 1} = 0

3sin^2(x)+11sin(x)+10=0

3 sin^{2} (x) + 11 sin (x) + 10 = 0

cos(x)(cos(x)+5)=2sin^2(x)

cos (x) (cos (x) + 5) = 2 sin^{2} (x)

Frequently Asked Questions (FAQ)

What is the general solution for 4sin(2t)=4cos(t) ?
The general solution for 4sin(2t)=4cos(t) is t= pi/2+2pin,t=(3pi)/2+2pin,t= pi/6+2pin,t=(5pi)/6+2pin