What is the general solution for sin(2x)=sqrt(2)cos(x),0<= x<= 2pi ?

Q: What is the general solution for sin(2x)=sqrt(2)cos(x),0<= x<= 2pi ?

The general solution for sin(2x)=sqrt(2)cos(x),0<= x<= 2pi is x= pi/2 ,x=(3pi)/2 ,x= pi/4 ,x=(3pi)/4

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sin(2x)=sqrt(2)cos(x),0<= x<= 2pi

Solution

sin (2 x) = 2 cos (x), 0 \leq x \leq 2 π

Solution

x = \frac{π}{2}, x = \frac{3 π}{2}, x = \frac{π}{4}, x = \frac{3 π}{4}

Degrees

x = 9 0^{\circ}, x = 27 0^{\circ}, x = 4 5^{\circ}, x = 13 5^{\circ}

Solution steps

sin (2 x) = 2 cos (x), 0 \leq x \leq 2 π

Subtract

2 cos (x)

from both sides

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

Rewrite using trig identities

sin (2 x) - cos (x) 2

Use the Double Angle identity:

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

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

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

Factor

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

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

Factor out common term

cos (x)

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

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

Solving each part separately

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

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

cos (x) = 0, 0 \leq x \leq 2 π

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

Solutions for the range

0 \leq x \leq 2 π

x = \frac{π}{2}, x = \frac{3 π}{2}

- 2 + 2 sin (x) = 0, 0 \leq x \leq 2 π : x = \frac{π}{4}, x = \frac{3 π}{4}

- 2 + 2 sin (x) = 0, 0 \leq x \leq 2 π

Move

2

to the right side

- 2 + 2 sin (x) = 0

Add

2

to both sides

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

Simplify

2 sin (x) = 2

2 sin (x) = 2

Divide both sides by

2

2 sin (x) = 2

Divide both sides by

2

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

Simplify

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

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

General solutions for

sin (x) = \frac{2}{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}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

Solutions for the range

0 \leq x \leq 2 π

x = \frac{π}{4}, x = \frac{3 π}{4}

Combine all the solutions

x = \frac{π}{2}, x = \frac{3 π}{2}, x = \frac{π}{4}, x = \frac{3 π}{4}

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tan(x)=(-1)/1

tan (x) = \frac{- 1}{1}

sqrt(4cos(θ))=1-cos(θ)

4 cos (θ) = 1 - cos (θ)

sin(x/3)=0

sin (\frac{x}{3}) = 0

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cos (x) = 0.95

Frequently Asked Questions (FAQ)

What is the general solution for sin(2x)=sqrt(2)cos(x),0<= x<= 2pi ?
The general solution for sin(2x)=sqrt(2)cos(x),0<= x<= 2pi is x= pi/2 ,x=(3pi)/2 ,x= pi/4 ,x=(3pi)/4