sinx+cosx=2 has how many solutions and how?
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Answered by
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sinx+cosx=2—(1)sinx+cosx=2—(1)
Divide by 2–√2on both LHSandRHS=>sinx∗12√+cosx∗12√=22√=2–√—(2)LHSandRHS=>sinx∗12+cosx∗12=22=2—(2)
Since sin45=cos45=12√sin45=cos45=12, the equation(2)(2)can be written assinxcos45+cosxsin45=2–√—(3)sinxcos45+cosxsin45=2—(3)
SincesinAcosB+cosAsinB=sin(A+B)sinAcosB+cosAsinB=sin(A+B), equation (3)(3)can be written as sin(x+45)=2–√—(4)sin(x+45)=2—(4)
The known facts are:
The value for sinAsinAis always between−1−1and112–√=1.4142=1.414, which is greater than 1
From the above facts, LHS of the equation (4) is always between -1 and 1 whereas RHS is > 1. This is not possible and hence six+cosxsix+cosxcan never be equal to 2.
Divide by 2–√2on both LHSandRHS=>sinx∗12√+cosx∗12√=22√=2–√—(2)LHSandRHS=>sinx∗12+cosx∗12=22=2—(2)
Since sin45=cos45=12√sin45=cos45=12, the equation(2)(2)can be written assinxcos45+cosxsin45=2–√—(3)sinxcos45+cosxsin45=2—(3)
SincesinAcosB+cosAsinB=sin(A+B)sinAcosB+cosAsinB=sin(A+B), equation (3)(3)can be written as sin(x+45)=2–√—(4)sin(x+45)=2—(4)
The known facts are:
The value for sinAsinAis always between−1−1and112–√=1.4142=1.414, which is greater than 1
From the above facts, LHS of the equation (4) is always between -1 and 1 whereas RHS is > 1. This is not possible and hence six+cosxsix+cosxcan never be equal to 2.
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Answered by
2
let take sinx = sin90
let take cosx=cos0
such that sin90=1
cos 0=1
according to problem
sinx+cosx= 2
sin90+cos0 =2
1+1 =2
hence it is proved
let take cosx=cos0
such that sin90=1
cos 0=1
according to problem
sinx+cosx= 2
sin90+cos0 =2
1+1 =2
hence it is proved
bro x value should be taken one
yes bro according to valves table
Take x value same if sinx=90 then Cosx must be 90 according to our problem..Probpem is Cosx+sinx=2
no
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