What is the general solution for cos(θ)=sqrt(3/4) ?

The general solution for cos(θ)=sqrt(3/4) is θ= pi/6+2pin,θ=(11pi)/6+2pin

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cos(θ)=sqrt(3/4)

cos (θ) = \frac{3}{4}

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

Solution steps

cos (θ) = \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}

cos (θ) = \frac{3}{2}

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

2 sec (4 x) - 2 = 0

A, y = - 6 cos (A) - 2 sin^{2} (A)

2 cos (x^{2}) - 1 = 0

sin^{4} (x) = cos^{4} (x)

10. 5^{2} = 6. 7^{2} + 7. 8^{2} - 2 \cdot 6.7 \cdot 7.8 \cdot cos (x)

What is the general solution for cos(θ)=sqrt(3/4) ?
The general solution for cos(θ)=sqrt(3/4) is θ= pi/6+2pin,θ=(11pi)/6+2pin