P1 June 2009 Q4-ii

[Mortal9090]

P1 June 2009 Q4-ii

find the smallest value of x in the interval 0<=x<=2pi for which y=0

it uses the already built equation of y=6sin(2x)+3

how to solve? pls help

 

[DragonCub]

Simply solve it as an algebraic equation:
6 sin(2x) + 3 = 0
sin(2x) = -1/2
2x = 2pi /3
x = pi/3

 

[Saturation]

See, you just form the equation when y = 0 :

6sin(2x)+3 = 0
6sin(2x) = -3
sin(2x)= -3/6 = -0.5

Then solve it like a normal trignometrical equation, and use the smallest value for x!

sin ( A ) = -0.5 2x = A therefore 0 =< A =< 12.57
A = sin inverse -0.5 = -0.523 ( which is out of the range for A )
pi - -0.523 = 3.665

2x = 3.665
Therefore, X = 1.83

 

[aaakhtar19]

hey if anyone have any queury regarding past
papers pls pm me or post here i hv solved all
pastpapers ....................

and answering ur queuries wud help me practice more thnx

 

[Mortal9090]

See, you just form the equation when y = 0 :

6sin(2x)+3 = 0
6sin(2x) = -3
sin(2x)= -3/6 = -0.5

Then solve it like a normal trignometrical equation, and use the smallest value for x!

sin ( A ) = -0.5 2x = A therefore 0 =< A =< 12.57
A = sin inverse -0.5 = -0.523 ( which is out of the range for A )
pi - -0.523 = 3.665

2x = 3.665
Therefore, X = 1.83

Your answer is right mate.

A few questions, where does that 12.57 comes from?

Also, I am seeing you did not use double angle formula and using some other method? (i.e. Sin2A)

 

[Saturation]

Actually using the range involving 12.57 is part of solving the trigonometrical equation! If you use another method to solve trigonometrical equations, stick to it! This is how I solve them. And btw, i haven't heard of the double angle formula, so yeah, if you use another method, use that, but the bottom line is that you need to solve the equation :

sin(2x)= -0.5

Anyhow, this is where the 12.57 comes from ( which is from the method I use):

X can lie between 0 =< x =< 2pi

2x = A

So, A lies between 2(0) =< A =< 2(2pi) (Substituting the maximum and minimum values of X) So this means that A can have a minimum of 0 and maximum of 12.57

 

[Mortal9090]

Actually using the range involving 12.57 is part of solving the trigonometrical equation! If you use another method to solve trigonometrical equations, stick to it! This is how I solve them. And btw, i haven't heard of the double angle formula, so yeah, if you use another method, use that, but the bottom line is that you need to solve the equation :

sin(2x)= -0.5

Anyhow, this is where the 12.57 comes from ( which is from the method I use):

X can lie between 0 =< x =< 2pi

2x = A

So, A lies between 2(0) =< A =< 2(2pi) (Substituting the maximum and minimum values of X) So this means that A can have a minimum of 0 and maximum of 12.57

oh yes 12.57~=4pi

that makes sense..

fyi double angle=http://mathworld.wolfram.com/Double-AngleFormulas.html