What is the general solution for 16cos(2θ)-9=0 ?

The general solution for 16cos(2θ)-9=0 is θ=(0.97338…)/2+pin,θ=pi-(0.97338…)/2+pin

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16cos(2θ)-9=0

16 cos (2 θ) - 9 = 0

θ = \frac{0.97338 \dots}{2} + πn, θ = π - \frac{0.97338 \dots}{2} + πn

Degrees

θ = 27.88556 \dots^{\circ} + 18 0^{\circ} n, θ = 152.11443 \dots^{\circ} + 18 0^{\circ} n

Solution steps

16 cos (2 θ) - 9 = 0

Move

9

to the right side

16 cos (2 θ) - 9 = 0

Add

9

to both sides

16 cos (2 θ) - 9 + 9 = 0 + 9

Simplify

16 cos (2 θ) = 9

16 cos (2 θ) = 9

Divide both sides by

16

16 cos (2 θ) = 9

Divide both sides by

16

\frac{16 cos ( 2 θ )}{16} = \frac{9}{16}

Simplify

cos (2 θ) = \frac{9}{16}

cos (2 θ) = \frac{9}{16}

Apply trig inverse properties

cos (2 θ) = \frac{9}{16}

General solutions for

cos (2 θ) = \frac{9}{16}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

2 θ = arccos (\frac{9}{16}) + 2 πn, 2 θ = 2 π - arccos (\frac{9}{16}) + 2 πn

2 θ = arccos (\frac{9}{16}) + 2 πn, 2 θ = 2 π - arccos (\frac{9}{16}) + 2 πn

Solve

2 θ = arccos (\frac{9}{16}) + 2 πn : θ = \frac{arccos ( \frac{9}{16} )}{2} + πn

2 θ = arccos (\frac{9}{16}) + 2 πn

Divide both sides by

2

2 θ = arccos (\frac{9}{16}) + 2 πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{arccos ( \frac{9}{16} )}{2} + \frac{2 πn}{2}

Simplify

θ = \frac{arccos ( \frac{9}{16} )}{2} + πn

θ = \frac{arccos ( \frac{9}{16} )}{2} + πn

Solve

2 θ = 2 π - arccos (\frac{9}{16}) + 2 πn : θ = π - \frac{arccos ( \frac{9}{16} )}{2} + πn

2 θ = 2 π - arccos (\frac{9}{16}) + 2 πn

Divide both sides by

2

2 θ = 2 π - arccos (\frac{9}{16}) + 2 πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{2 π}{2} - \frac{arccos ( \frac{9}{16} )}{2} + \frac{2 πn}{2}

Simplify

θ = π - \frac{arccos ( \frac{9}{16} )}{2} + πn

θ = π - \frac{arccos ( \frac{9}{16} )}{2} + πn

θ = \frac{arccos ( \frac{9}{16} )}{2} + πn, θ = π - \frac{arccos ( \frac{9}{16} )}{2} + πn

Show solutions in decimal form

θ = \frac{0.97338 \dots}{2} + πn, θ = π - \frac{0.97338 \dots}{2} + πn

0 = 2 + 3 cos (θ) + cos (2 θ)

sin (2 θ) = cos (θ), 2

\frac{1 + cos ^{2} ( x )}{sin ^{2} ( x )} = 1

\frac{sin ^{2} ( θ ) - 2 sin ( θ ) + 1}{sin ( θ ) - 1} = - 1

w, 100 cos (10 w) = 50

What is the general solution for 16cos(2θ)-9=0 ?
The general solution for 16cos(2θ)-9=0 is θ=(0.97338…)/2+pin,θ=pi-(0.97338…)/2+pin