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    How many real solutions are there to tan x + (1/ tan x) = 0?
    The answer is zero because tan squared x = -1 has no solution which I can't understand.
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    (Original post by TheAdviser101)
    How many real solutions are there to tan x + (1/ tan x) = 0?
    The answer is zero because tan squared x = -1 has no solution which I can't understand.
    \tan x + \frac{1}{\tan x} = 0

    \tan^2 x + 1 = 0

    \tan^2 x = -1

    For real values of x, LHS is positive, whereas RHS is negative. Hence, there are no real solutions.
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    (Original post by SherlockHolmes)
    \tan x + \frac{1}{\tan x} = 0

    \tan^2 x + 1 = 0

    \tan^2 x = -1

    For real values of x, LHS is positive, whereas RHS is negative. Hence, there are no real solutions.
    Ohh thanks Sherlock 😀
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    (Original post by SherlockHolmes)
    \tan x + \frac{1}{\tan x} = 0

    \tan^2 x + 1 = 0

    \tan^2 x = -1

    For real values of x, LHS is positive, whereas RHS is negative. Hence, there are no real solutions.
    Sorry to bother you again but can you help me with the this question:
    Solve 3sin(x)(squared) + sin (x) -2 =0. 0<x<2pi
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    1/tan(x)(tan^2(x)+1) = 0 is the format for this way of factorising. When you see a 1/tanx 1/sinx And 1/Cosx or any functions of x thèse are factoriseable in the above way a lot of the Time. Then the solutions are tanx = plus or minus Root 1
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    (Original post by TheAdviser101)
    Sorry to bother you again but can you help me with the this question:
    Solve 3sin(x)(squared) + sin (x) -2 =0. 0<x<2pi
    Notice that 3\sin^2 x + \sin x -2 = 0 is a quadratic in \sin x.
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    (Original post by SherlockHolmes)
    Notice that 3\sin^2 x + \sin x -2 = 0 is a quadratic in \sin x.
    So it factorises to (3sinx -2)(sinx +1)
    Giving the solutions 0.73, 2.4 and (3/2)pi ??
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    (Original post by TheAdviser101)
    So it factorises to (3sinx -2)(sinx +1)
    Giving the solutions 0.73, 2.4 and (3/2)pi ??
    Yes, that's right. Except I think it should be 2.41 if to two decimal places.
 
 
 
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