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Identities question help

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(edited 6 years ago)
Reply 1
Avatar for ThomH97
ThomH97
21
For the first question, what are the two values you have?

For the second, can you make the equation into a quadratic in cos(x)?
Original post by eatingkiwi
Solve 2sin^2x = 5cosx + 4 between 0 and 360 degrees.
I have no idea how to go about this one.

Please could somebody explain what to do?


three
For the first one you should have gotten a quadratic with 2 roots. That will give you 4 answers by the end.

For the second one you want sin2x+cos2x=1sin^2x+cos^2x=1
Reply 4
Avatar for dguess499
dguess499
Solve 2sin^2x = 5cosx + 4 between 0 and 360 degrees. I have no idea how to go about this one.
sin^2x+cos^2x=1

2(1-cos^2x) = 5cosx + 4
if X=cosx
then -2X^2-5X-2=0
delta=25-4*-2*-2=3^2
then X1=... and X2=... then x=....
Reply 5
Avatar for Zacken
Zacken
22
Original post by dguess499
Solve 2sin^2x = 5cosx + 4 between 0 and 360 degrees. I have no idea how to go about this one.
sin^2x+cos^2x=1

2(1-cos^2x) = 5cosx + 4
if X=cosx
then -2X^2-5X-2=0
delta=25-4*-2*-2=3^2
then X1=... and X2=... then x=....


Can you solve the quadratic -2X^2 - 5X -2 = 0?

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