[koat]

hello
can somebody help me with the question
solve the equation sin2x+cos2x=0 for (-pi/2<_x<_pi/2)
i only get -pi/8 and 7/8pi which is wrong.
answer says 3/8 pi
can somebody help please
thanks

[xiyangliu]

(Original post by koat)
hello
can somebody help me with the question in the mei book p. 217 question 10i
i only get -pi/8 and 7/8pi which is wrong.
can somebody help please
thanks

Let me take a look


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[raheem94]

(Original post by koat)
hello
can somebody help me with the question in the mei book p. 217 question 10i
i only get -pi/8 and 7/8pi which is wrong.
can somebody help please
thanks

It will be better if you post the question, everyone won't have the mei book.

[xiyangliu]

U should get -1/8pi and 3/8 pi.

Do u knw how I got there ??


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[xiyangliu]

Divide the equation by cos2x

So u get tan2x = -1

Solve for 2x = -1/4pi an 3/4pi

Then divide these two by 2. So Now all solution between the limit is found


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[Amy-Rose]

(Original post by koat)
how do you get 3/4pi ?

tanx=-1

2x=-1/4pi, using CAST shows tangent is positive in third quadrant, so we do pi+-1/4pi to get 3/4pi

therefore half these,

x=-1/8pi, 3/8pi ---> both of which are in the range.

[koat]

(Original post by xiyangliu)
Let me take a look


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thanks
you know i dont quite get part v
because i get the x values -0.366 and 1.366 in part iv but the values in part i are smaller so why does the book say in the solutions angle not small