need help in trig?

Question

(cos(t))^2=1/2 find solutions between 0 and 360.  I got pi/4 and 7pi/4. Pls help idk how im wrong and explain to me how to do pls.

husoski · Answer

"Between 0 and 360" implies degrees, but you are working in radians.

Either way, there should be 4 answers, since the cosine could be either 1/sqrt(2) or -1/sqrt(2) before squaring to get 1/2.  Once you know that pi/4 (aka 45 degrees) is a solution, the other three are easy

t1 = 45
t2 = 360 - 45 = 315 .... cost t2 = 1/sqrt(2)
t3 = 180 - 45 = 135 .... cos t3 = -1/sqrt(2)
t4 = 180 + 45 = 225 .... cos t4 = -1/sqrt(5)

Robert · Answer

Since (cos t)^2 = 1/2, cos t = ± √2 / 2.

That adds 3π/4 and 5π/4 to the answer.

MechEng2030 · Answer

You forgot to consider the negative case.

cos(t) = +/- sqrt(2)/2
t = pi/4, 3pi/4, 5pi/4, 7pi/4

? · Answer

cosx = ± sqrt(2)/2!!

Kareem · Answer

cos (t)^2= 1/2
cos (t)= 1/sqrt(2)
cos (t)= sqrt(2)/2 (PLUS or MINUS because it is Square root)
t= (pi/4, 3pi/4, 5pi/4, 7pi/4)

need help in trig?

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