Math, asked by takailala, 9 months ago

How do I solve cos2x(2cosx+1)=0

Answers

Answered by dandaniaub2012
0

Answer:s=(1/2,π/3)

Step-by-step explanation:

cos2x(2cosx+1)=0

cos2x=0 or 2cosx +1 = 0

2x = cos0 and cosx = 1/2

x = cos0/2 and x = cos1/2

x = 1/2 and x = π/3

Answered by faithturrell14
0

Answer:

Explanation: Solving each part separately, cos(2x)=0or2cos(x)+1=0. Now, cos(2x)=0,0≤x≤2π Again,we have, 2cos(x)+1=0,0≤x≤2π cos(x)=−12: x=2π3+2πn, x=4π3+2πn.

Step-by-step explanation:

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