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

The general solution for 2cos(x)+cos(2x)=0 is x=1.19606…+2pin,x=2pi-1.19606…+2pin

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

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

x = 1.19606 \dots + 2 πn, x = 2 π - 1.19606 \dots + 2 πn

Solution steps

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

Rewrite using trig identities

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

Solve by substitution

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

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

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

No Solution

Combine all the solutions

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

Show solutions in decimal form

x = 1.19606 \dots + 2 πn, x = 2 π - 1.19606 \dots + 2 πn

x, - sin (x + y) = 0

cot (θ) = - \frac{1}{2}

sin (2 θ) + sin (θ) + 2 cos (θ) + 1 = 0, 0 \leq θ \leq 2 π

sin^{2} (t) = cos^{2} (t)

cos (x) - sin (x) sin (2 x) = 0

What is the general solution for 2cos(x)+cos(2x)=0 ?
The general solution for 2cos(x)+cos(2x)=0 is x=1.19606…+2pin,x=2pi-1.19606…+2pin