What is the general solution for 4cos^2(θ)-3=0,0<= θ<= 2pi ?

Q: What is the general solution for 4cos^2(θ)-3=0,0<= θ<= 2pi ?

The general solution for 4cos^2(θ)-3=0,0<= θ<= 2pi is θ= pi/6 ,θ=(11pi)/6 ,θ=(5pi)/6 ,θ=(7pi)/6

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4cos^2(θ)-3=0,0<= θ<= 2pi

Solution

4 cos^{2} (θ) - 3 = 0, 0 \leq θ \leq 2 π

Solution

θ = \frac{π}{6}, θ = \frac{11 π}{6}, θ = \frac{5 π}{6}, θ = \frac{7 π}{6}

Degrees

θ = 3 0^{\circ}, θ = 33 0^{\circ}, θ = 15 0^{\circ}, θ = 21 0^{\circ}

Solution steps

4 cos^{2} (θ) - 3 = 0, 0 \leq θ \leq 2 π

Solve by substitution

4 cos^{2} (θ) - 3 = 0

Let:

cos (θ) = u

4 u^{2} - 3 = 0

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

4 u^{2} - 3 = 0

Move

3

to the right side

4 u^{2} - 3 = 0

Add

3

to both sides

4 u^{2} - 3 + 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 = cos (θ)

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

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

cos (θ) = \frac{3}{2}, 0 \leq θ \leq 2 π : θ = \frac{π}{6}, θ = \frac{11 π}{6}

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

General solutions for

cos (θ) = \frac{3}{2}

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}

θ = \frac{π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

θ = \frac{π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

Solutions for the range

0 \leq θ \leq 2 π

θ = \frac{π}{6}, θ = \frac{11 π}{6}

cos (θ) = - \frac{3}{2}, 0 \leq θ \leq 2 π : θ = \frac{5 π}{6}, θ = \frac{7 π}{6}

cos (θ) = - \frac{3}{2}, 0 \leq θ \leq 2 π

General solutions for

cos (θ) = - \frac{3}{2}

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}

θ = \frac{5 π}{6} + 2 πn, θ = \frac{7 π}{6} + 2 πn

θ = \frac{5 π}{6} + 2 πn, θ = \frac{7 π}{6} + 2 πn

Solutions for the range

0 \leq θ \leq 2 π

θ = \frac{5 π}{6}, θ = \frac{7 π}{6}

Combine all the solutions

θ = \frac{π}{6}, θ = \frac{11 π}{6}, θ = \frac{5 π}{6}, θ = \frac{7 π}{6}

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Frequently Asked Questions (FAQ)

What is the general solution for 4cos^2(θ)-3=0,0<= θ<= 2pi ?
The general solution for 4cos^2(θ)-3=0,0<= θ<= 2pi is θ= pi/6 ,θ=(11pi)/6 ,θ=(5pi)/6 ,θ=(7pi)/6