How many solutions does this equation have?
(4 sinX -√5) (sinX+1)
Here is a link to the sqa marking instructions. It is question 11. Can someone explain it in greater depth? Thanks
http://www.sqa.org.uk/pastpapers/papers/instructions/2009/mi_H_Mathematics_all_2009.pdf
(4 sinX -√5) (sinX+1)
Here is a link to the sqa marking instructions. It is question 11. Can someone explain it in greater depth? Thanks
http://www.sqa.org.uk/pastpapers/papers/instructions/2009/mi_H_Mathematics_all_2009.pdf
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(4 sin X -√5 ) ( sin X+1 ) = 0
4 sin X = √5 , sin X = - 1
sin X = √5 / 4 , sin X = - 1
X = 34 ° , 146 ° , 270 °
4 sin X = √5 , sin X = - 1
sin X = √5 / 4 , sin X = - 1
X = 34 ° , 146 ° , 270 °
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Thank you for rating----pleased to help.
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Either you are missing " = 0 " at the end (to make it an equation), or the question asks - maybe implicitly - to find the roots of the expression
(a root is a value given to the variable, that will cause the expression to add up to zero).
Here, you have two factors. It ANY one of the factors is zero, then the product will be zero.
Each one has a cyclical function, with a finite period (the period of the sine function is 2*pi, meaning that sin(x) = sin(x + 2*pi) = sin(x + 4*pi) = ... = sin(x + 2k*pi) where k is any integer (positive or negative).
Therefore, each factor can have an infinite number of solutions (because there are an infinite number of integers available for k).
However, in such problems, it is customary to restrict the domain (the possible input value - the value of x) to a finite interval, usually covering only one cycle, for example:
Domain = [0, 2*pi)
So, if we restrict ourselves to that domain:
The second factor (the easy one) will be equal to zero ONLY when sin(x) = -1, and this happens only at x = 3*pi/2
The first factor will be equal to zero ONLY when sin(x) = + (√5)/4
and this happens twice per cycle.
Therefore, the expression has three roots per cycle.
From what I can see, the questions does not ask for the actual value of the roots, only the number of roots.
(a root is a value given to the variable, that will cause the expression to add up to zero).
Here, you have two factors. It ANY one of the factors is zero, then the product will be zero.
Each one has a cyclical function, with a finite period (the period of the sine function is 2*pi, meaning that sin(x) = sin(x + 2*pi) = sin(x + 4*pi) = ... = sin(x + 2k*pi) where k is any integer (positive or negative).
Therefore, each factor can have an infinite number of solutions (because there are an infinite number of integers available for k).
However, in such problems, it is customary to restrict the domain (the possible input value - the value of x) to a finite interval, usually covering only one cycle, for example:
Domain = [0, 2*pi)
So, if we restrict ourselves to that domain:
The second factor (the easy one) will be equal to zero ONLY when sin(x) = -1, and this happens only at x = 3*pi/2
The first factor will be equal to zero ONLY when sin(x) = + (√5)/4
and this happens twice per cycle.
Therefore, the expression has three roots per cycle.
From what I can see, the questions does not ask for the actual value of the roots, only the number of roots.