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\[solve ~for ~~~~0 \le x \le 2x\]\[5\cot x +2 +2\cot^2x=5\]
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2cot^2(x)+5cot(x)-3=0 (2cotx-1)(cotx+3)=0 cotx=1/2, cotx=-3 tanx=2, tanx=-1/3 x=63.4, 243.4,161.6, 341.6 (Degrees) x=1.11, 4.249, 2.821, 5.963 (Radians)
|dw:1331040651608:dw|
x=63.4, 243.4,161.6, 341.6 (Degrees) x=1.11, 4.249, 2.821, 5.963 (Radians) How did you get this? In both degrees and radians?
click you calculator \[x=\tan^{-1} 2 \] For that x value, because the value "2" is positive (just look at the positive region), look at the thing that i draw for "tangent" |dw:1331040989326:dw| So,just now the value for \[x=\tan^{-1} 2 \] x=63.43, then use that 180+63.43= 243.43 and get the "next" x value so, you got x=63.43, 243.43
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