SOLUTION: Hello everyone, and Merry Christmas!!
Can somebody explain me the difference about these two kind of solutions
starting from following equation:
sin(x)cos(x) + cos^2(x) = 0
a)
Algebra ->
Trigonometry-basics
-> SOLUTION: Hello everyone, and Merry Christmas!!
Can somebody explain me the difference about these two kind of solutions
starting from following equation:
sin(x)cos(x) + cos^2(x) = 0
a)
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Question 1062390: Hello everyone, and Merry Christmas!!
Can somebody explain me the difference about these two kind of solutions
starting from following equation:
sin(x)cos(x) + cos^2(x) = 0
a) first method dividing by cos^2(x) I become tan(x) + 1 = 0 wich gives me
the result: tan(x) = -1; x= -PI/4 + nPI; n€Z
b) second method dividing by sin^2(x) I become cot(x) + cot^2(x) = 0 wich gives
me the two results: cot(x) = -1; x= -PI/4 + nPI; n€Z
cot(x) = 0; x= PI/2 + nPI; n€Z
Where am I doing or thinking wrong? Thank you very much RB
You can put this solution on YOUR website! sin(x)cos(x) + cos^2(x) = 0
a) first method dividing by cos^2(x) I become tan(x) + 1 = 0 wich gives me
the result: tan(x) = -1; x= -PI/4 + nPI; n€Z
b) second method dividing by sin^2(x) I become cot(x) + cot^2(x) = 0 wich gives
me the two results: cot(x) = -1; x= -PI/4 + nPI; n€Z
cot(x) = 0; x= PI/2 + nPI; n€Z
Where am I doing or thinking wrong? Thank you very much RB
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sin(x)cos(x) + cos^2(x) = 0
cos*(sin + cos) = 0
cos = 0
x = pi/2 + n*pi, n = any integer
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sin + cos = 0
tan + 1 = 0
tan = -1
x = 3pi/4, 7pi/4 + n*pi, n = any integer
