What is the general solution for 4cos^2(x)+cos(2x)-7cos(x)=-2 ?

The general solution for 4cos^2(x)+cos(2x)-7cos(x)=-2 is x=2pin,x=1.40334…+2pin,x=2pi-1.40334…+2pin

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4cos^2(x)+cos(2x)-7cos(x)=-2

4 cos^{2} (x) + cos (2 x) - 7 cos (x) = - 2

x = 2 πn, x = 1.40334 \dots + 2 πn, x = 2 π - 1.40334 \dots + 2 πn

Solution steps

4 cos^{2} (x) + cos (2 x) - 7 cos (x) = - 2

Subtract

- 2

from both sides

4 cos^{2} (x) + cos (2 x) - 7 cos (x) + 2 = 0

Rewrite using trig identities

1 + 6 cos^{2} (x) - 7 cos (x) = 0

Solve by substitution

cos (x) = 1, cos (x) = \frac{1}{6}

cos (x) = 1 : x = 2 πn

cos (x) = \frac{1}{6} : x = arccos (\frac{1}{6}) + 2 πn, x = 2 π - arccos (\frac{1}{6}) + 2 πn

Combine all the solutions

x = 2 πn, x = arccos (\frac{1}{6}) + 2 πn, x = 2 π - arccos (\frac{1}{6}) + 2 πn

Show solutions in decimal form

x = 2 πn, x = 1.40334 \dots + 2 πn, x = 2 π - 1.40334 \dots + 2 πn

4 cos (x) = 2 + 2 cos (x)

4 cos (θ) = 43 sin (θ)

- 4 sin (θ) = 4 (sin (θ) - 1)

4 + 2 cos (θ) = 0

9.81 sin (θ) - 5.88 cos (θ) = 2.77

What is the general solution for 4cos^2(x)+cos(2x)-7cos(x)=-2 ?
The general solution for 4cos^2(x)+cos(2x)-7cos(x)=-2 is x=2pin,x=1.40334…+2pin,x=2pi-1.40334…+2pin