What is the general solution for 8sin(2x)=8cos(x) ?

The general solution for 8sin(2x)=8cos(x) is x= pi/2+2pin,x=(3pi)/2+2pin,x= pi/6+2pin,x=(5pi)/6+2pin

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8sin(2x)=8cos(x)

Solution

8 sin (2 x) = 8 cos (x)

Solution

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

Degrees

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

Solution steps

8 sin (2 x) = 8 cos (x)

Subtract

8 cos (x)

from both sides

8 sin (2 x) - 8 cos (x) = 0

Rewrite using trig identities

- 8 cos (x) + 8 sin (2 x)

Use the Double Angle identity:

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

= - 8 cos (x) + 8 \cdot 2 sin (x) cos (x)

Simplify

= - 8 cos (x) + 16 sin (x) cos (x)

- 8 cos (x) + 16 cos (x) sin (x) = 0

Factor

- 8 cos (x) + 16 cos (x) sin (x) : 8 cos (x) (2 sin (x) - 1)

- 8 cos (x) + 16 cos (x) sin (x)

Rewrite

16

2 \cdot 8

= - 8 cos (x) + 2 \cdot 8 sin (x) cos (x)

Factor out common term

8 cos (x)

= 8 cos (x) (- 1 + 2 sin (x))

8 cos (x) (2 sin (x) - 1) = 0

Solving each part separately

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

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

cos (x) = 0

General solutions for

cos (x) = 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}

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

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

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

2 sin (x) - 1 = 0

Move

1

to the right side

2 sin (x) - 1 = 0

Add

1

to both sides

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

Simplify

2 sin (x) = 1

2 sin (x) = 1

Divide both sides by

2

2 sin (x) = 1

Divide both sides by

2

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

Simplify

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

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

General solutions for

sin (x) = \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}

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

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

Combine all the solutions

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

Popular Examples

cos(t)=(sqrt(2))/2

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

tan^2(θ)+5tan(θ)+6=0

tan^{2} (θ) + 5 tan (θ) + 6 = 0

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

2 cos^{2} (x) - sin^{2} (x) = 1

2cos^2(θ)-sin(θ)-1=0

2 cos^{2} (θ) - sin (θ) - 1 = 0

sin(t)=-1

sin (t) = - 1

Frequently Asked Questions (FAQ)

What is the general solution for 8sin(2x)=8cos(x) ?
The general solution for 8sin(2x)=8cos(x) is x= pi/2+2pin,x=(3pi)/2+2pin,x= pi/6+2pin,x=(5pi)/6+2pin