What is the general solution for 2cos(2θ)-3cos(θ)-2=0 ?

The general solution for 2cos(2θ)-3cos(θ)-2=0 is θ=2.33643…+2pin,θ=-2.33643…+2pin

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2cos(2θ)-3cos(θ)-2=0

2 cos (2 θ) - 3 cos (θ) - 2 = 0

θ = 2.33643 \dots + 2 πn, θ = - 2.33643 \dots + 2 πn

Solution steps

2 cos (2 θ) - 3 cos (θ) - 2 = 0

Rewrite using trig identities

- 4 - 3 cos (θ) + 4 cos^{2} (θ) = 0

Solve by substitution

cos (θ) = \frac{3 + 73}{8}, cos (θ) = \frac{3 - 73}{8}

cos (θ) = \frac{3 + 73}{8} :

No Solution

cos (θ) = \frac{3 - 73}{8} : θ = arccos (\frac{3 - 73}{8}) + 2 πn, θ = - arccos (\frac{3 - 73}{8}) + 2 πn

Combine all the solutions

θ = arccos (\frac{3 - 73}{8}) + 2 πn, θ = - arccos (\frac{3 - 73}{8}) + 2 πn

Show solutions in decimal form

θ = 2.33643 \dots + 2 πn, θ = - 2.33643 \dots + 2 πn

tan (2 θ) + 6 tan (θ) + 8 = 0

cos (φ) = 3 sin (φ)

5 sin (x) - 3 cos (2 x) + 3 = 0

\frac{tan ^{2} ( x )}{sec ( x ) + 1} = tan (x)

2 cos^{2} (x) + cos (x) = 1, 0 \leq x < 2 π

What is the general solution for 2cos(2θ)-3cos(θ)-2=0 ?
The general solution for 2cos(2θ)-3cos(θ)-2=0 is θ=2.33643…+2pin,θ=-2.33643…+2pin