6cos^2-5sin-5=0

Question

can someone please help me find Q1 2 3 4

Answer 1

6cos2A – 5sinA – 5 = 0
6(cos2A) – 5sinA – 5 = 0
6(1-sin2A) – 5sinA – 5 = 0     [ recall: cos2A + sin2A = 1 ]
6 – 6sin²A - 5sinA – 5 = 0
0 = -6 + 6sin²A + 5sinA + 5
0 = 6sin²A + 5sinA – 1
0 = 6(sinA)² + 5(sinA) – 1     [now factor trinomial…]
0 = (6sinA – 1)(sinA + 1)

(6sinA – 1)=0                  -OR-      (sinA + 1)=0
   6sinA = 1                     -OR-        sinA = -1
 A = sin-1(1/6)                -OR-       A = sin-1(-1)
 A = 9.5451°, 170.41°      -OR-      A = 270°

 Thus, A = { 9.5451°, 170.41°, 270° }

Answer 2

We want everything to have a consistent trig function.  We can get everything in terms of sine.

 

6(1 - sin²(x)) - 5sin(x) - 5 = 0

 

6 - 6sin²(x) - 5sin(x) - 5=

 

-6sin²(x) - 5sin(x) + 1 = 0

 

 

We can use the quadratic formula to find sin(x).

 

sin(x) = (5 ± √(25 - 4(-6))) / -12

 

sin(x) = (5 ± √(25 + 24) / -12

 

sin(x) = (5 ± √(49) / -12

 

sin(x) = (5 ± 7) / -12

 

sin(x) = (5 + 7) / -12            and          sin(x) = (5 - 7) / -12

 

sin(x) = 12 / -12                  and           sin(x) = -2 / -12

 

sin(x) = -1                            and           sin(x) = 1/6

 

x = 270                             and               x = 9.6

                                                              x = 180 - 9.60 = 170.40    (2nd quadrant)

 

There are 3 solutions:

 

x = 9.60

x = 170.40

x = 270

Answer 3

6Cos²θ-5Sinθ-5=0

 

Remember: Cos²θ+Sin²θ=1

 

6(1-Sin²θ)-5Sinθ-5=0

 

6-6Sin2θ-5Sinθ-5=0

 

6Sin²θ+5Sinθ-1=0

 

Now you have a quadratic equation that you need to solve for Sinθ. The root are:

 

Sinθ=-1 => θ=(Π/2)(4n+3), n∈Z

Sinθ=1/6 => θ=2Πn+Sin^-1(1/6), n∈Z