Trigonometry Equation

6 cos

^{2 }0 = 3Please I need step by step instruction on how the answers where found.

Thank You for being so Helpfull,

Bella

Trigonometry Equation

6 cos ^{2 }0 = 3

Please I need step by step instruction on how the answers where found.

Thank You for being so Helpfull,

Bella

Tutors, please sign in to answer this question.

Decatur, AL

6 cos^{2} x = 3

Is that your question?

If so,

Divide both sides by 6:

(6 cos^{2} x)/6 = 3/6

cos^{2} x = 1/2

Take the square root of both sides:

cos x = +/- sqrt(1/2)

cos x = +/- (√2)/2

Cos x = +/- Sqrt(2)/2 at every multiple of pi/4

cos pi/4 = Sqrt(2)/2

cos 3pi/4 = - sqrt(2)/2

cos 5pi/4 = - sqrt(2)/2

cos 7pi/4 = sqrt(2)/2

Is that your question?

If so,

Divide both sides by 6:

(6 cos

cos

Take the square root of both sides:

cos x = +/- sqrt(1/2)

cos x = +/- (√2)/2

Cos x = +/- Sqrt(2)/2 at every multiple of pi/4

cos pi/4 = Sqrt(2)/2

cos 3pi/4 = - sqrt(2)/2

cos 5pi/4 = - sqrt(2)/2

cos 7pi/4 = sqrt(2)/2

...

So x = (pi)(2n+1)/4 for n = 0,1,2,3...

It depends on if they gave you an interval to use for the solution set.

For example, they could have said x or "theta" such that 0<= x <= 2pi.

If so, the answers would be x = {pi/4, 3pi/4, 5pi/4, 7pi/4}

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