2sinxcosx - sin(2x)cos(2x) = 0
<span>Part I </span>
<span>The double angle identity for sine states that sin(2x) = 2sinxcosx </span>
<span>Thus we get: </span>
<span>sin(2x) - sin(2x)cos(2x) = 0 </span>
<span>Part II </span>
<span>sin(2x)(1 - cos(2x)) = 0 </span>
<span>Part III </span>
<span>Either sin(2x) = 0 or </span>
<span>1 - cos(2x) = 0 </span>
<span>=> cos(2x) = 1 </span>
<span>For sin(2x) = 0, this is true for </span>
<span>2x = n(pi) where n = 0, 1, 2, .... </span>
<span>x = n(pi/2) </span>
<span>For cos(2x) = 1, this is true for </span>
<span>2x = n(pi) where n = 0, 2, 4, .... </span>
<span>x = n(pi/2)
</span>
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Answer:
He multiplied the radius by 2 instead of squaring it
One dollar and eighty cents for wayside hotel
Two dollars and seventy five cents for blue sky hotel
(4,4).........................
