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2sec^2x - 3tanx - 5
2sec^2x -3tanx =5 for [0,2pi)
2 answers
asked by
tom
923 views
2sec^2x - 3tanx - 5 = 0
over the interval 0 to 2pi x= 1.14, 4.28, 5.68, and 2.54 right?
2 answers
asked by
tom
965 views
2sec^2x -3tanx =5 for [0,2pi)
Please solve for the values of x for me.
2 answers
asked by
tom
479 views
2sec^2x - 3tanx - 5 = 0
I got x= 1.24 radians x= 4.38 radians x= -0.74 radians x= -3.88 radians
2 answers
asked by
SkiMasktheSlumpGod
426 views
Could someone please help with this question.
State the properties of y = 2sec(-2x + 180°) + 3. My answer: First I graphed y =
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Am I right on this?
2sec^2(2x)-8=0 2sec^2(2x)=8 divide by 2 sec^2(2x)=4 find square root sec(2x)=2 Now, would this simply be
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