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

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

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Popular >

sin(2t)=cos(t)

Solution

sin (2 t) = 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

sin (2 t) = cos (t)

Subtract

cos (t)

from both sides

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

Rewrite using trig identities

- cos (t) + sin (2 t)

Use the Double Angle identity:

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

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

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

Factor

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

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

Factor out common term

cos (t)

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

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

cos(2x)+1=cos(x)

cos (2 x) + 1 = cos (x)

4cos(x)+1=0

4 cos (x) + 1 = 0

2cot^2(x)+csc^2(x)-2=0

2 cot^{2} (x) + csc^{2} (x) - 2 = 0

cos^2(x)= 1/2 sin(2x)

cos^{2} (x) = \frac{1}{2} sin (2 x)

3cot(θ)+5=0

3 cot (θ) + 5 = 0

Frequently Asked Questions (FAQ)

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