# What is a complicated math problem that equals 3?

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# What’s a sophisticated math downside that equals 3?

### 7 Solutions

• Restrict instance.

lim x→∞ (3x² + x)/(x² + 1) = 3

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• Math Equations That Equal 3

• Not difficult, however considerably superior:

Discover the primary by-product of f(x) = 3x^4 – 4x^3 + x^2 + x + 5 when x = 1

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If sin2x=22 – 8√7 Show that 3(Sinx+Cosx+Tanx+Cosecx+Secx+Cotx) =21 Let sin 2x = ok ok = 22- 8√7 ok – 22 = -8√7 [Bring 22 to the LHS) (k-22)² = (-8√7)² [Squaring both sides] k² – 44k + 484 = 448 [Expanding] k² – 44k = 448-484 [Bringing 484 to the RHS] k² – 44k = -36 [Multiply throughout by k] k³ – 44k² = -36k k³ – 44k² + 36k = 0 [Bringing -36k to the LHS] k³ – 49k² +5k² + 8k + 28k = 0 [Rearranging factors] k³ + 5k² + 8k = 49k² – 28k [Rearranging factors and bring 49k² – 28k to the RHS] k³ + 4k² + k² + 4k + 4k = 49k² – 28k [Rearranging factors] k³ + 4k² + k² + 4k + 4k + 4 – 4 = 49k² – 28k [Adding + 4 and -4 to make it a quadratic/polynomial] k³ + 4k² + k² + 4k + 4k + 4 = 49k² – 28k + 4 [Bringing -4 to the RHS] k³ + 4k² + k² + 4k + 4k + 4 = (7k – 2)² [Writing 49k² – 28k + 4 in the form of (a-b)²] (1+ok)(k² + 4k + 4) = (7k – 2)² [Factoring LHS] (1+ok)(ok + 2)² = (7k – 2)² [Writing k² + 4k + 4 in the form of (a+b)²] (1+ok) = (7k-2)²/(ok+2)² [Bringing (k+2)² to the RHS] (1 + 2sinxcosx) = (7k-2)²/(ok+2)² [k = sin2x = 2sinxcosx] (sin²x + cos²x + 2sinxcosx) = (7k-2)²/(ok+2)² [ 1 = sin²x + cos²x and we bring LHS to the form (a+b)²] (sinx + cosx)² = (7k-2)²/(ok+2)² (sinx + cosx) = (7k-2)/(ok+2) [Taking root on both sides] sinx + cosx = (7k -2)/ok / (ok+2)/ok) [Divide Numerator and Denominator of RHS by k] sinx + cosx = (7 – 2/ok) / (1+2/ok) (sinx + cosx) (1 + 2/ok) = 7 – 2/ok [Taking (1+2/k) to the LHS] sinx + cosx + 2sinx/ok + 2cosx/ok = 7 – 2/ok [Multiplying and Expanding] sinx + cosx + 2/ok(sinx + cosx) + 2/ok = 7 [Taking Common factor out] sinx + cosx + 2/ok(sinx + cosx + 1) = 7 [Taking Common factor out] sinx + cosx + 2/ok(sinx + cosx + sin²x + cos²x) = 7 [ 1 = sin²x + cos²x] sinx + cosx + 2/2sinxcos (sinx + cosx + sin²x + cos²x) = 7 [k = sin2x = 2sinxcosx] sinx + cosx + 1/cosx + 1/sinx + sinx/cosx + cosx/sinx = 7 [Canceling common terms] sinx+cosx+secx+cosecx+tanx+cotx = 7 [1/cosx=secx,1/sinx=cosecx,sinx/cosx=tan… Therefore sinx + cosx + secx + cosecx + tanx + cotx = 7 3(sinx + cosx + secx + cosecx + tanx + cotx) = 21 Cheers !!!

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