Question

Solve sin2x + 5sinx + 1 + 5cosx = 0​

Answer 1

Given: sin2x + 5sinx + 1 + 5cosx = 0

To find: We have to solve this.

Solution:

sin2x + 5sinx + 1 + 5cosx = 0

We know that-

${sin}^{2} x + {cos}^{2} x = 1$

Putting this value of 1 in the above equation we get-

$sin2x + 5sinx + 1 + 5cosx = 0 \\ sin2x + 5sinx + {sin}^{2}x + {cos}^{2} x + 5cosx = 0$

Again we know that Sin2x=2sinxcosx

Putting the value of sin2x in the above equation we get-

$2sinx.cosx + 5sinx + {sin}^{2}x + {cos}^{2} x + 5cosx = 0 \\ 2sinx.cosx + 5sinx + {(sinx + cosx)}^{2} - 2sinx.cosx + 5cosx = 0 \\ 5sinx + {(sinx + cosx)}^{2} + 5cosx = 0 \\ 5(sinx + cosx) + {(sinx + cosx)}^{2} = 0 \\ (sinx + cosx)(sinx + cosx + 5) = 0$

So, we get two factors -

$(sinx + cosx) = 0 \\ sinx = - cosx \\ tanx = - 1$

$(sinx + cosx + 5) = 0 \\ sinx + cosx = - 5$

Answer 2

x = nπ - π/4  if sin2x + 5sinx + 1 + 5cosx = 0 where n ∈Z

Given :

• sin2x + 5sinx + 1 + 5cosx = 0

To Find:

• Solve for x

Solution:

Step 1:

Use identity  sin²x + cos²x = 1  and  sin 2x = 2sinxcosx and rearrange terms

2sinxcosx + 5sinx + sin²x + cos²x + 5cosx = 0

sin²x + cos²x +  2sinxcosx+ 5sinx + 5cosx = 0

Step 2:

Use identity  (a + b)² = a² + b² + 2ab  and take (sinx + cosx) common factor

(sinx + cosx)² +  5(sinx + cosx) = 0

(sinx + cosx)(sinx + cosx + 5) =0

Step 3:

Equate each factor with zero and solve for x

sinx + cosx + 5 = 0 is not possible  as min value of sinx + cosx is -√2

sinx + cosx = 0

=> sinx = -cosx

=> tanx = - 1

x = nπ - π/4  where n ∈Z

x = nπ - π/4  if sin2x + 5sinx + 1 + 5cosx = 0  where n ∈Z