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C3-trig

hi can someone solve this
3secsquaredX-4=0

can someone help me

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Reply 1
Avatar for steve2005
steve2005
17
Original post by otrivine
hi can someone solve this
3secsquaredX-4=0

can someone help me


You could write sec in terms of cos
Reply 2
Avatar for raheem94
raheem94
13
Original post by otrivine
hi can someone solve this
3secsquaredX-4=0

can someone help me


3sec2x−4=0  ⟹  3sec2x=4 \displaystyle 3sec^2x-4=0 \implies 3sec^2x=4

Remember, sec2x=1cos2x \displaystyle sec^2x = \frac1{cos^2x}
Reply 3
Avatar for r_t
r_t
3/cos^2 (x) = 4
cos^2(x) = 3/4
cos (x) = root3/2
x = pi/6 or 30 degrees
Reply 4
Avatar for raheem94
raheem94
13
Original post by r_t
3/cos^2 (x) = 4
cos^2(x) = 3/4
cos (x) = root3/2
x = pi/6 or 30 degrees


It will be better to give OP some hints, so that he does it himself, rather than post a full solution for him.

By the way, your solution isn't completely correct.
cos2x=34  ⟹  cosx=±32 \displaystyle cos^2x=\frac34 \implies cosx= \pm \frac{\sqrt3}{2}

You haven't considered the negative case.
Reply 5
Avatar for otrivine
otrivine
OP
14
Original post by raheem94
3sec2x−4=0  ⟹  3sec2x=4 \displaystyle 3sec^2x-4=0 \implies 3sec^2x=4

Remember, sec2x=1cos2x \displaystyle sec^2x = \frac1{cos^2x}


i get it so secX=3/4 and you invert it and change the root to +4/3 then do cos-1(root3/2)
(edited 11 years ago)
Reply 6
Avatar for steve2005
steve2005
17
Original post by otrivine
i get it so secX=3/4 and you invert it and change the root to +4/3 then do cos-1(root3/2)


secX does not equal 3/4
Reply 7
Avatar for raheem94
raheem94
13
Original post by otrivine
i get it so secX=3/4 and you invert it and change the root to +4/3 then do cos-1(root3/2)


I am not understanding what you are trying to do.

3sec2x−4=0  ⟹  3sec2x=4  ⟹  sec2x=43  ⟹  1cos2x=43  ⟹  cos2x=34 \displaystyle 3sec^2x-4=0 \implies 3sec^2x=4 \implies sec^2x=\frac43 \implies \frac1{cos^2x} = \frac43 \\ \implies cos^2x = \frac34

Do you get it?
Reply 8
Avatar for otrivine
otrivine
OP
14
Original post by raheem94
I am not understanding what you are trying to do.

3sec2x−4=0  ⟹  3sec2x=4  ⟹  sec2x=43  ⟹  1cos2x=43  ⟹  cos2x=34 \displaystyle 3sec^2x-4=0 \implies 3sec^2x=4 \implies sec^2x=\frac43 \implies \frac1{cos^2x} = \frac43 \\ \implies cos^2x = \frac34

Do you get it?


ohhh yes yes i get it now :wink:
and one more thing for this question tansquared2X-2sec2x+1=0
i got 90 degrees and one maths error is this correct so far or not?
as for the equation i got secsquared2x-2sec2x+1=0
Reply 9
Avatar for raheem94
raheem94
13
Original post by otrivine
ohhh yes yes i get it now :wink:
and one more thing for this question tansquared2X-2sec2x+1=0
i got 90 degrees and one maths error is this correct so far or not?
as for the equation i got secsquared2x-2sec2x+1=0


It isn't correct.

tan22x−2sec2x+1=0 tan^22x-2sec2x+1=0

Remember, tan22x=sec22x−1 tan^22x=sec^22x-1

So the equation becomes,
tan22x−2sec2x+1=0  ⟹  sec22x−1−2sec2x+1=0  ⟹  sec22x−2sec2x=0 tan^22x-2sec2x+1=0 \implies sec^22x-1-2sec2x+1=0 \\ \implies sec^22x-2sec2x=0
Reply 10
Avatar for otrivine
otrivine
OP
14
Original post by raheem94
It isn't correct.

tan22x−2sec2x+1=0 tan^22x-2sec2x+1=0

Remember, tan22x=sec22x−1 tan^22x=sec^22x-1

So the equation becomes,
tan22x−2sec2x+1=0  ⟹  sec22x−1−2sec2x+1=0  ⟹  sec22x−2sec2x=0 tan^22x-2sec2x+1=0 \implies sec^22x-1-2sec2x+1=0 \\ \implies sec^22x-2sec2x=0


yes that is what i got but my answer are wrong :s-smilie:

i applied quadratic formula X2-2X+0 which gives 2 and 0
Original post by otrivine
yes that is what i got but my answer are wrong :s-smilie:

i applied quadratic formula X2-2X+0 which gives 2 and 0


Factorise
Reply 12
Avatar for raheem94
raheem94
13
Original post by otrivine
yes that is what i got but my answer are wrong :s-smilie:

i applied quadratic formula X2-2X+0 which gives 2 and 0


You don't need to apply the quadratic formula,

sec22x−2sec2x=0  ⟹  sec2x(sec2x−2)=0 \displaystyle sec^22x-2sec2x=0 \implies sec2x(sec2x-2)=0

So you get 2 equations, sec2x=0 and sec2x−2=0 sec2x=0 \text{ and } sec2x-2 =0
Reply 13
Avatar for otrivine
otrivine
OP
14
Original post by steve2005
Factorise


yes i get same answer 0 , 2
Reply 14
Avatar for otrivine
otrivine
OP
14
Original post by raheem94
You don't need to apply the quadratic formula,

sec22x−2sec2x=0  ⟹  sec2x(sec2x−2)=0 \displaystyle sec^22x-2sec2x=0 \implies sec2x(sec2x-2)=0

So you get 2 equations, sec2x=0 and sec2x−2=0 sec2x=0 \text{ and } sec2x-2 =0


i get 90 degrees and maths error

and then i takeaway 90 with 360 then keep dividing by 2 correct?

cause my book i giving me only 2 solution 0,180
Reply 15
Avatar for raheem94
raheem94
13
Original post by otrivine
yes i get same answer 0 , 2


See my previous post.

0 and 2 are correct because the 2 equation are sec2x=0 and sec2x=2.

But you need to use sec2x=1cos2x \displaystyle sec2x= \frac1{cos2x} to find the values of '2x' and hence 'x'
Reply 16
Avatar for raheem94
raheem94
13
Original post by otrivine
i get 90 degrees and maths error

and then i takeaway 90 with 360 then keep dividing by 2 correct?

cause my book i giving me only 2 solution 0,180


Are the answers in your book 0 and 180?
Reply 17
Avatar for otrivine
otrivine
OP
14
Original post by raheem94
See my previous post.

0 and 2 are correct because the 2 equation are sec2x=0 and sec2x=2.

But you need to use sec2x=1cos2x \displaystyle sec2x= \frac1{cos2x} to find the values of '2x' and hence 'x'


ok so dont we have to do firstly cos-1(0) then divide by 2 ?
Reply 18
Avatar for otrivine
otrivine
OP
14
Original post by raheem94
Are the answers in your book 0 and 180?


yes its the edexcel c3 text book
Original post by otrivine
i get 90 degrees and maths error

and then i takeaway 90 with 360 then keep dividing by 2 correct?

cause my book i giving me only 2 solution 0,180


I think you are looking at the wrong answers.

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