What is the general solution for cos(2θ)+6sin^2(θ)=4,0<= x<2pi ?

Q: What is the general solution for cos(2θ)+6sin^2(θ)=4,0<= x<2pi ?

The general solution for cos(2θ)+6sin^2(θ)=4,0<= x<2pi is θ= pi/3 ,θ=(2pi)/3 ,θ=(4pi)/3 ,θ=(5pi)/3

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cos(2θ)+6sin^2(θ)=4,0<= x<2pi

Solution

cos (2 θ) + 6 sin^{2} (θ) = 4, 0 \leq x < 2 π

Solution

θ = \frac{π}{3}, θ = \frac{2 π}{3}, θ = \frac{4 π}{3}, θ = \frac{5 π}{3}

Degrees

θ = 6 0^{\circ}, θ = 12 0^{\circ}, θ = 24 0^{\circ}, θ = 30 0^{\circ}

Solution steps

cos (2 θ) + 6 sin^{2} (θ) = 4, 0 \leq x < 2 π

Subtract

4

from both sides

cos (2 θ) + 6 sin^{2} (θ) - 4 = 0

Rewrite using trig identities

- 4 + cos (2 θ) + 6 sin^{2} (θ)

Use the Double Angle identity:

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

= - 4 + 1 - 2 sin^{2} (θ) + 6 sin^{2} (θ)

Simplify

- 4 + 1 - 2 sin^{2} (θ) + 6 sin^{2} (θ) : 4 sin^{2} (θ) - 3

- 4 + 1 - 2 sin^{2} (θ) + 6 sin^{2} (θ)

Add similar elements:

- 2 sin^{2} (θ) + 6 sin^{2} (θ) = 4 sin^{2} (θ)

= - 4 + 1 + 4 sin^{2} (θ)

Add/Subtract the numbers:

- 4 + 1 = - 3

= 4 sin^{2} (θ) - 3

= 4 sin^{2} (θ) - 3

- 3 + 4 sin^{2} (θ) = 0

Solve by substitution

- 3 + 4 sin^{2} (θ) = 0

Let:

sin (θ) = u

- 3 + 4 u^{2} = 0

- 3 + 4 u^{2} = 0 : u = \frac{3}{2}, u = - \frac{3}{2}

- 3 + 4 u^{2} = 0

Move

3

to the right side

- 3 + 4 u^{2} = 0

Add

3

to both sides

- 3 + 4 u^{2} + 3 = 0 + 3

Simplify

4 u^{2} = 3

4 u^{2} = 3

Divide both sides by

4

4 u^{2} = 3

Divide both sides by

4

\frac{4 u ^{2}}{4} = \frac{3}{4}

Simplify

u^{2} = \frac{3}{4}

u^{2} = \frac{3}{4}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = \frac{3}{4}, u = - \frac{3}{4}

\frac{3}{4} = \frac{3}{2}

\frac{3}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{3}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{3}{2}

- \frac{3}{4} = - \frac{3}{2}

- \frac{3}{4}

Simplify

\frac{3}{4} : \frac{3}{2}

\frac{3}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{3}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{3}{2}

= - \frac{3}{2}

u = \frac{3}{2}, u = - \frac{3}{2}

Substitute back

u = sin (θ)

sin (θ) = \frac{3}{2}, sin (θ) = - \frac{3}{2}

sin (θ) = \frac{3}{2}, sin (θ) = - \frac{3}{2}

sin (θ) = \frac{3}{2}, 0 \leq x < 2 π : θ = \frac{π}{3}, θ = \frac{2 π}{3}

sin (θ) = \frac{3}{2}, 0 \leq x < 2 π

General solutions for

sin (θ) = \frac{3}{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}

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

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

Solutions for the range

0 \leq x < 2 π

θ = \frac{π}{3}, θ = \frac{2 π}{3}

sin (θ) = - \frac{3}{2}, 0 \leq x < 2 π : θ = \frac{4 π}{3}, θ = \frac{5 π}{3}

sin (θ) = - \frac{3}{2}, 0 \leq x < 2 π

General solutions for

sin (θ) = - \frac{3}{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}

θ = \frac{4 π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

θ = \frac{4 π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

Solutions for the range

0 \leq x < 2 π

θ = \frac{4 π}{3}, θ = \frac{5 π}{3}

Combine all the solutions

θ = \frac{π}{3}, θ = \frac{2 π}{3}, θ = \frac{4 π}{3}, θ = \frac{5 π}{3}

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Frequently Asked Questions (FAQ)

What is the general solution for cos(2θ)+6sin^2(θ)=4,0<= x<2pi ?
The general solution for cos(2θ)+6sin^2(θ)=4,0<= x<2pi is θ= pi/3 ,θ=(2pi)/3 ,θ=(4pi)/3 ,θ=(5pi)/3