SOLUTION: Solve: 9sin2x-28cos2x=0, 0°<x≤90°.

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Question 1203570: Solve: 9sin2x-28cos2x=0, 0°
Found 2 solutions by Theo, greenestamps:
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
equation is 9 * sin(2x) - 28 * cos(2x) = 0
add 28 * cos(2x) to both sides of the equation to get:
9 * sin(2x) = 28 * cos(2x)
divide both sides of the equation by 28 * cos(2x) to get:
9 * sin(2x) / 28 * cos(2x) = 1
since sin/cos = tan, you get:
9 / 28 * tan(2x) = 1
multiply both sides of the equation by 28/9 to get:
tan(2x) = 28/9
solve for 2x to get:
2x = 72.18111109 degrees.
replace 2x in the original equation to get:
9 * sin(72.18111109) - 28 * cos(72.18111109) = 0
evaluate using your calculator to get:
0 = 0, confirming the value of 2x is good.
divide 2x by 2 to get:
x = 36.09055554 degrees.
that's your angle.

graphically, this will look like this:



your solution will be when 2x = 72.181111 degrees and when 2x = 252.181111 degrees.

that means that x = 36.091 degrees or 126.091 degrees.

when x is 36.0191, both sin and cos are positive.
when x is 126.091, both sin and cos are negative.

you were asked to find the value of x between 0 and 90 degrees.
that would be x = 36.091 degrees.

note that, when x = 36.091, y = 9 * sin (2 * 36.091) = 9 * sin(72.181).
values displayed on the graph are rounded to 3 decimal places.



Answer by greenestamps(13215) About Me  (Show Source):
You can put this solution on YOUR website!


Parts of your equation are the expressions "sin2x" and "cos2x".

Are those sin%282x%29 and cos%282x%29?

Or are they %28sinx%29%5E2 and cos%28x%29%5E2?