What is the general solution for cos(x+pi/6)cos(x-pi/6)=cos(2x) ?

The general solution for cos(x+pi/6)cos(x-pi/6)=cos(2x) is x=(5pi)/6+pin,x= pi/6+pin

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cos(x+pi/6)cos(x-pi/6)=cos(2x)

cos (x + \frac{π}{6}) cos (x - \frac{π}{6}) = cos (2 x)

x = \frac{5 π}{6} + πn, x = \frac{π}{6} + πn

Solution steps

cos (x + \frac{π}{6}) cos (x - \frac{π}{6}) = cos (2 x)

Rewrite using trig identities

(\frac{3}{2} cos (x) - \frac{1}{2} sin (x)) (\frac{3}{2} cos (x) + \frac{1}{2} sin (x)) = cos (2 x)

Subtract

cos (2 x)

from both sides

(\frac{3}{2} cos (x) - \frac{1}{2} sin (x)) (\frac{3}{2} cos (x) + \frac{1}{2} sin (x)) - cos (2 x) = 0

Simplify

(\frac{3}{2} cos (x) - \frac{1}{2} sin (x)) (\frac{3}{2} cos (x) + \frac{1}{2} sin (x)) - cos (2 x) : \frac{( 3 cos ( x ) - sin ( x ) ) ( 3 cos ( x ) + sin ( x ) ) - 4 cos ( 2 x )}{4}

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

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

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

Rewrite using trig identities

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

Factor

- cos^{2} (x) + 3 sin^{2} (x) : (3 sin (x) + cos (x)) (3 sin (x) - cos (x))

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

Solving each part separately

3 sin (x) + cos (x) = 0 or 3 sin (x) - cos (x) = 0

3 sin (x) + cos (x) = 0 : x = \frac{5 π}{6} + πn

3 sin (x) - cos (x) = 0 : x = \frac{π}{6} + πn

Combine all the solutions

x = \frac{5 π}{6} + πn, x = \frac{π}{6} + πn

arctan (\frac{x}{12}) - arctan (x) = - π

θ, z \cdot p \cdot cos (θ) = 5

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

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

θ, z \cdot p \cdot cos (θ) = n

What is the general solution for cos(x+pi/6)cos(x-pi/6)=cos(2x) ?
The general solution for cos(x+pi/6)cos(x-pi/6)=cos(2x) is x=(5pi)/6+pin,x= pi/6+pin