What is the general solution for 2cos(9x)=2-2cos(9x) ?

The general solution for 2cos(9x)=2-2cos(9x) is x= pi/(27)+(2pin)/9 ,x=(5pi}{27}+\frac{2pin)/9

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2cos(9x)=2-2cos(9x)

2 cos (9 x) = 2 - 2 cos (9 x)

x = \frac{π}{27} + \frac{2 πn}{9}, x = \frac{5 π}{27} + \frac{2 πn}{9}

Solution steps

2 cos (9 x) = 2 - 2 cos (9 x)

Solve by substitution

cos (9 x) = \frac{1}{2}

cos (9 x) = \frac{1}{2} : x = \frac{π}{27} + \frac{2 πn}{9}, x = \frac{5 π}{27} + \frac{2 πn}{9}

Combine all the solutions

x = \frac{π}{27} + \frac{2 πn}{9}, x = \frac{5 π}{27} + \frac{2 πn}{9}

sin (2 x) = 32, - π \leq x \leq π

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

cos^{2} (x) + sec^{2} (x) = 11

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

csc (2 x + 3^{\circ}) = \frac{5}{4}

What is the general solution for 2cos(9x)=2-2cos(9x) ?
The general solution for 2cos(9x)=2-2cos(9x) is x= pi/(27)+(2pin)/9 ,x=(5pi}{27}+\frac{2pin)/9