What is the general solution for cos^2(x)+cos^4(x)+cos^6(x)=0 ?

The general solution for cos^2(x)+cos^4(x)+cos^6(x)=0 is x= pi/2+2pin,x=(3pi)/2+2pin

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cos^2(x)+cos^4(x)+cos^6(x)=0

cos^{2} (x) + cos^{4} (x) + cos^{6} (x) = 0

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Solution steps

cos^{2} (x) + cos^{4} (x) + cos^{6} (x) = 0

Solve by substitution

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

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = \frac{1}{2} + \frac{3}{2} i :

No Solution

cos (x) = - \frac{1}{2} - \frac{3}{2} i :

No Solution

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

No Solution

cos (x) = \frac{1}{2} - \frac{3}{2} i :

No Solution

Combine all the solutions

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

sin (x) = \frac{- 1}{4}

sin^{4} (x) - sin^{2} (x) = 0

(cos (t) - 4) (2 sin^{2} (t) - 1) = 0

sin (7 5^{\circ}) = \frac{x}{9}

i, 2 cos^{3} (x) + sin (x) + 1 = 2 sin^{2} (x)

What is the general solution for cos^2(x)+cos^4(x)+cos^6(x)=0 ?
The general solution for cos^2(x)+cos^4(x)+cos^6(x)=0 is x= pi/2+2pin,x=(3pi)/2+2pin