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if 2 cosx + sinx = 1 then what is the value of 7cosx + 6sinx
answer is 1
Cosx = (1-tan2x/2)/(1+tan2x/2) Sin x = (2tanx/2)/(1+tan2x/2) 2cosx + sinx=1 2(((1-tan2x/2)/(1+tan2x/2)) + ((2tanx/2)/(1+tan2x/2))=1 Solving this we get quadratic in tanx/2 3tan2x/2 – 2tanx/2 -1=0 (3tanx/2+1)(tanx/2 -1)=0 Tanx/2 =1 or -1/3 7cosx + 6sinx 7((1-tan2x/2)/(1+tan2x/2)) + 6((2tanx/2)/(1+tan2x/2)) Putting value of Tanx/2, we get When Tanx/2 =1 Ans=6 When Tanx/2=-1/3 Ans=2
Cosx = (1-tan2x/2)/(1+tan2x/2)
Sin x = (2tanx/2)/(1+tan2x/2)
2cosx + sinx=1
2(((1-tan2x/2)/(1+tan2x/2)) + ((2tanx/2)/(1+tan2x/2))=1
Solving this we get quadratic in tanx/2
3tan2x/2 – 2tanx/2 -1=0
(3tanx/2+1)(tanx/2 -1)=0
Tanx/2 =1 or -1/3
7cosx + 6sinx
7((1-tan2x/2)/(1+tan2x/2)) + 6((2tanx/2)/(1+tan2x/2))
Putting value of Tanx/2, we get
When
Tanx/2 =1 Ans=6
Tanx/2=-1/3 Ans=2
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