Finding roots of this trig eqn

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The equation [sinx] + [√2 cosx] = -3 requires both terms to be at their minimum values, which occurs when sinx and cosx are at specific angles. The left-hand side achieves a minimum of -2, indicating that the equation cannot be satisfied as -3 is unattainable. The valid intervals for x based on the greatest integer function are narrowed down to either (5π/4, 2π) or (π, 5π/4). Further analysis shows that checking values one by one is not the most efficient method for solving this equation. The conclusion is that the equation has no solutions within the specified range.
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Homework Statement


If [sinx]+[√2 cosx]=-3, x belongs to [0,2∏] (where [.] denotes the greatest integer function) then x belongs to
1)[5∏/4,2∏]
2)(5∏/4,2∏)
3)(∏,5∏/4)
4)[∏,5∏/4]

The Attempt at a Solution



If I put x=5∏/4, LHS=-2 which does not satisfy. 2∏ and ∏ also does not satisfy.So the answer must be either 2) or 3). Is there any better way other than putting the values one by one and checking?
 
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utkarshakash said:

Homework Statement


If [sinx]+[√2 cosx]=-3, x belongs to [0,2∏] (where [.] denotes the greatest integer function) then x belongs to
1)[5∏/4,2∏]
2)(5∏/4,2∏)
3)(∏,5∏/4)
4)[∏,5∏/4]

The Attempt at a Solution



If I put x=5∏/4, LHS=-2 which does not satisfy. 2∏ and ∏ also does not satisfy.So the answer must be either 2) or 3). Is there any better way other than putting the values one by one and checking?
What are the lowest possible values of the two terms on the left? How can they add to -3?
 
haruspex said:
What are the lowest possible values of the two terms on the left? How can they add to -3?

That was helpful. Thanks!
 
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