What is the general solution for 4cos(2x)+3cos(x)=1 ?

The general solution for 4cos(2x)+3cos(x)=1 is x=0.89566…+2pin,x=2pi-0.89566…+2pin,x=pi+2pin

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4cos(2x)+3cos(x)=1

4 cos (2 x) + 3 cos (x) = 1

x = 0.89566 \dots + 2 πn, x = 2 π - 0.89566 \dots + 2 πn, x = π + 2 πn

Solution steps

4 cos (2 x) + 3 cos (x) = 1

Subtract

1

from both sides

4 cos (2 x) + 3 cos (x) - 1 = 0

Rewrite using trig identities

- 5 + 3 cos (x) + 8 cos^{2} (x) = 0

Solve by substitution

cos (x) = \frac{5}{8}, cos (x) = - 1

cos (x) = \frac{5}{8} : x = arccos (\frac{5}{8}) + 2 πn, x = 2 π - arccos (\frac{5}{8}) + 2 πn

cos (x) = - 1 : x = π + 2 πn

Combine all the solutions

x = arccos (\frac{5}{8}) + 2 πn, x = 2 π - arccos (\frac{5}{8}) + 2 πn, x = π + 2 πn

Show solutions in decimal form

x = 0.89566 \dots + 2 πn, x = 2 π - 0.89566 \dots + 2 πn, x = π + 2 πn

0 = 2 sin (x)

cos (x) = \frac{2}{3}

4 sin (θ) + 5 = 7

6 arcsin (x) = π

sin (x) = \frac{- 1}{2}

What is the general solution for 4cos(2x)+3cos(x)=1 ?
The general solution for 4cos(2x)+3cos(x)=1 is x=0.89566…+2pin,x=2pi-0.89566…+2pin,x=pi+2pin