What is the general solution for 1/(cos^2(θ/2))=12cos(θ) ?

The general solution for 1/(cos^2(θ/2))=12cos(θ) is θ=1.42478…+2pin,θ=2pi-1.42478…+2pin

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1/(cos^2(θ/2))=12cos(θ)

\frac{1}{cos ^{2} ( \frac{θ}{2} )} = 12 cos (θ)

θ = 1.42478 \dots + 2 πn, θ = 2 π - 1.42478 \dots + 2 πn

Solution steps

\frac{1}{cos ^{2} ( \frac{θ}{2} )} = 12 cos (θ)

Subtract

12 cos (θ)

from both sides

\frac{1}{cos ^{2} ( \frac{θ}{2} )} - 12 cos (θ) = 0

Simplify

\frac{1}{cos ^{2} ( \frac{θ}{2} )} - 12 cos (θ) : \frac{1 - 12 cos ^{2} ( \frac{θ}{2} ) cos ( θ )}{cos ^{2} ( \frac{θ}{2} )}

\frac{1 - 12 cos ^{2} ( \frac{θ}{2} ) cos ( θ )}{cos ^{2} ( \frac{θ}{2} )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

1 - 12 cos^{2} (\frac{θ}{2}) cos (θ) = 0

Rewrite using trig identities

1 - (1 + cos (θ)) \cdot 6 cos (θ) = 0

Solve by substitution

cos (θ) = - \frac{3 + 15}{6}, cos (θ) = \frac{15 - 3}{6}

cos (θ) = - \frac{3 + 15}{6} :

No Solution

cos (θ) = \frac{15 - 3}{6} : θ = arccos (\frac{15 - 3}{6}) + 2 πn, θ = 2 π - arccos (\frac{15 - 3}{6}) + 2 πn

Combine all the solutions

θ = arccos (\frac{15 - 3}{6}) + 2 πn, θ = 2 π - arccos (\frac{15 - 3}{6}) + 2 πn

Show solutions in decimal form

θ = 1.42478 \dots + 2 πn, θ = 2 π - 1.42478 \dots + 2 πn

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

cos (x) = \frac{34}{70}

x, z = cos (x^{2} + 2 y)

- 3 sin (t) + 4 cos (t) = 0

cos (θ) = \frac{8}{6}

What is the general solution for 1/(cos^2(θ/2))=12cos(θ) ?
The general solution for 1/(cos^2(θ/2))=12cos(θ) is θ=1.42478…+2pin,θ=2pi-1.42478…+2pin