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

The general solution for 2cos(2t)+2cos^2(t)=0 is t=-0.95531…+pin,t=0.95531…+pin

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2cos(2t)+2cos^2(t)=0

2 cos (2 t) + 2 cos^{2} (t) = 0

t = - 0.95531 \dots + πn, t = 0.95531 \dots + πn

Solution steps

2 cos (2 t) + 2 cos^{2} (t) = 0

Rewrite using trig identities

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

Factor

- 2 sin^{2} (t) + 4 cos^{2} (t) : 2 (2 cos (t) + sin (t)) (2 cos (t) - sin (t))

2 (2 cos (t) + sin (t)) (2 cos (t) - sin (t)) = 0

Solving each part separately

2 cos (t) + sin (t) = 0 or 2 cos (t) - sin (t) = 0

2 cos (t) + sin (t) = 0 : t = arctan (- 2) + πn

2 cos (t) - sin (t) = 0 : t = arctan (2) + πn

Combine all the solutions

t = arctan (- 2) + πn, t = arctan (2) + πn

Show solutions in decimal form

t = - 0.95531 \dots + πn, t = 0.95531 \dots + πn

5 \cdot cos (x) + 4 - 2 \cdot sin (x) = 0

sin (x) = - 0.54

tan (2 θ) = 1.82

sin (x) = - 0.65

sin (x) = - 0.71

What is the general solution for 2cos(2t)+2cos^2(t)=0 ?
The general solution for 2cos(2t)+2cos^2(t)=0 is t=-0.95531…+pin,t=0.95531…+pin