sin ( x ) + cos ( x ) + tan ( x ) + csc ( x ) + sec ( x ) + cot ( x ) = 7
Suppose a real number x satisfy the equation above.
And for positive constants a , b , c , we have sin ( 2 x ) = a − b c with c square-free. Find the value of a − b + 2 c .
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Yes, we just need to group them such that we can apply the substitution sin ( 2 A ) = 2 sin ( A ) cos ( A ) .
Bonus question: With the same given equation, can you write a least degree polynomial such that it has a root of tan ( 2 x ) ?
To make your work neater, you can factor it nicely as:
7 sin ( 2 x ) − 2 = ( sin x + cos x ) ( 2 + sin ( 2 x ) )
Now, if we consider A = sin ( 2 x ) and square both sides of that equation we have, we get, after a bit of simplification,
A 3 − 4 4 A 2 + 3 6 A = 0 ⟹ A = 0 , 2 2 ± 8 7
We can easily rule out the two extraneous solutions A = 0 and A = 2 2 + 8 7 using the reason you provided. We get the answer as A = 2 2 − 8 7 .
I used the substitution sin ( x ) + cos ( x ) = A
So, sin ( x ) ⋅ cos ( x ) = 2 A 2 − 1
-> Find A
-> Evaluate A 2 − 1
Didn't think of solvable to sin 2 x which I suppose to realize as asked. Nevertheless, a general way is still workable. I think you made your own a solvable question. Nice to know this equation.
x = 1.07764686747605+
Sin 2 x = 0.83398951148327+ = 22 - 8 Sqrt (7)
22 - 8 + 2 (7) = 28
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