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

Solution

(1 + cos (x)) (1 + cos (2 x)) = \frac{1}{4}

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn, x = 1.25663 \dots + 2 πn, x = 2 π - 1.25663 \dots + 2 πn, x = 2.51327 \dots + 2 πn, x = - 2.51327 \dots + 2 πn

Solution steps

(1 + cos (x)) (1 + cos (2 x)) = \frac{1}{4}

Subtract

\frac{1}{4}

from both sides

cos (2 x) + cos (x) + cos (x) cos (2 x) + \frac{3}{4} = 0

Simplify

cos (2 x) + cos (x) + cos (x) cos (2 x) + \frac{3}{4} : \frac{4 cos ( 2 x ) + 4 cos ( x ) + 4 cos ( x ) cos ( 2 x ) + 3}{4}

\frac{4 cos ( 2 x ) + 4 cos ( x ) + 4 cos ( x ) cos ( 2 x ) + 3}{4} = 0

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

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

Rewrite using trig identities

- 1 + 8 cos^{2} (x) + 8 cos^{3} (x) = 0

Solve by substitution

cos (x) = - \frac{1}{2}, cos (x) = \frac{- 1 + 5}{4}, cos (x) = - \frac{1 + 5}{4}

cos (x) = - \frac{1}{2} : x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

cos (x) = \frac{- 1 + 5}{4} : x = arccos (\frac{- 1 + 5}{4}) + 2 πn, x = 2 π - arccos (\frac{- 1 + 5}{4}) + 2 πn

cos (x) = - \frac{1 + 5}{4} : x = arccos (- \frac{1 + 5}{4}) + 2 πn, x = - arccos (- \frac{1 + 5}{4}) + 2 πn

Combine all the solutions

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn, x = arccos (\frac{- 1 + 5}{4}) + 2 πn, x = 2 π - arccos (\frac{- 1 + 5}{4}) + 2 πn, x = arccos (- \frac{1 + 5}{4}) + 2 πn, x = - arccos (- \frac{1 + 5}{4}) + 2 πn

Show solutions in decimal form

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn, x = 1.25663 \dots + 2 πn, x = 2 π - 1.25663 \dots + 2 πn, x = 2.51327 \dots + 2 πn, x = - 2.51327 \dots + 2 πn

x, y + 3 = 2 sin (4 x)

tan^{4} (x) - 2 sec^{2} (x) + 3 = 0

sin (x) = 2 sin (x) - 1

tan (x) = - \frac{5}{3}

3 sin (θ) + 4 cos (θ) = 3