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# P2 trig watch

1. solve sin(x + pi/6) =2cosx

giving all solutions in teh range of 0 _< x _< 2pi
2. (Original post by hihihihi)
solve sin(x + pi/6) =2cosx

giving all solutions in teh range of 0 _< x _< 2pi
x=tan-1(sqrt3/2)
0.714 and 3.86
3. (Original post by lgs98jonee)
x=tan-1(sqrt3/2)
0.714 and 3.86
it says answer is pi/3, 4pi/3
4. (Original post by hihihihi)
it says answer is pi/3, 4pi/3
yeah sorry it x=tan-1(sqrt3)
=pi/3 or 4pi/3
5. sorry could you show workings pleeeeaaase?

How do you solve 5cos x + 12sinx=6 ?? I can't think of any identities or rules that will help, my brain is shot and its probably a simple trig q
7. (Original post by MJ_23)

How do you solve 5cos x + 12sinx=6 ?? I can't think of any identities or rules that will help, my brain is shot and its probably a simple trig q
Rsin(x+a)=R(sinxcosa+sinacosx)
equate coefficiants

Rsina=5
Rcosa=12

so tana=5/12
so a=22.6

so R=5/sin22.6
=13
13sin(x+22.6)=6
so x+22.6=arcsin(6/13)
=27.5 or 152.5 (between 0 and 360 degrees)
so x=4.9 or 129.9
8. (Original post by lgs98jonee)
Rsin(x+a)=R(sinxcosa+sinacosx)
equate coefficiants

Rsina=5
Rcosa=12

so tana=5/12
so a=22.6

so R=5/sin22.6
=13
13sin(x+22.6)=6
so x+22.6=arcsin(6/13)
=27.5 or 152.5 (between 0 and 360 degrees)
so x=175.1 or 50.1

Im not so sure.
9. (Original post by hihihihi)
solve sin(x + pi/6) =2cosx

giving all solutions in teh range of 0 _< x _< 2pi
sin(x+pi/6)=sinxcos(pi/6)+cosxsin(pi/6)
=sqrt3sinx/2+cosx/2.....[cos(pi/6)=sqrt3/2 and sin(pi/6)=0.5]
so sqrt3sinx/2=3cosx/2

divide both sides by cosx

so sqrt3tanx/2=3/2
so tanx=3/2*2/sqrt3
=3/sqrt3
=sqrt3
therefore x=pi/3 or 4pi/3
10. (Original post by Ralfskini)
Im not so sure.
yeah corrected mistake ( i had put x+22.6=blah and then put x=blah+22.6 rather than x=blah -22.6)
11. (Original post by lgs98jonee)
Rsin(x+a)=R(sinxcosa+sinacosx)
equate coefficiants

Rsina=5
Rcosa=12

so tana=5/12
so a=22.6

so R=5/sin22.6
=13
13sin(x+22.6)=6
so x+22.6=arcsin(6/13)
=27.5 or 152.5 (between 0 and 360 degrees)
so x=4.9 or 129.9
Yeah thats the answer in the book, thanks a lot.
12. (Original post by lgs98jonee)
sin(x+pi/6)=sinxcos(pi/6)+cosxsin(pi/6)
=sqrt3sinx/2+cosx/2.....[cos(pi/6)=sqrt3/2 and sin(pi/6)=0.5]
so sqrt3sinx/2=3cosx/2

divide both sides by cosx

so sqrt3tanx/2=3/2
so tanx=3/2*2/sqrt3
=3/sqrt3
=sqrt3
therefore x=pi/3 or 4pi/3
thanks v much got it now

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