Which of the following is a solution to 2cos^2x– 2 = 0 ?

a.0°
b.30°
c.180°
d.330°

A and C

Step-by-step explanation:

Given

2cos²x - 2 = 0 ( add 2 to both sides )

2cos²x = 2 ( divide both sides by 2 )

cos²x = 1 ( take the square root of both sides )

cos x = ± = ± 1

cos x = 1 ⇒ x = (1) = 0°

cos x  = - 1 ⇒ x = (- 1) = 180°

2 possible results

0° and 180°

Step-by-step explanation:

2cos² x– 2 = 0  (add 2 to both sides)

2cos² x = 2  (divide both sides by 2)

cos² x = 2/2

cos² x = 1

cos x = ±√1

cos x = ±1

because the range of x is not stated, there are 2 possible results for x  that will cause cos x to be either 1 or -1

0° and 180°

your answer is a

step-by-step explanation:

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step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?