solve the equation over the indicated domain 2cos^2x=3sinx=0 domain (0,2pi]

Question

solve the equation over the indicated domain 2cos^2x=3sinx=0 domain (0,2pi]

Answer 1

Since both trig terms equal the same value, we can equate the 2 functions.

 

2cos²(x) = 3sin(x)

 

 

Subtract 3sin(x) on both sides of equation.

 

2cos²(x) - 3sin(x) = 0

 

 

Use the identity  sin²(x) + cos²(x) = 1  to write the equation in terms of sine.

 

2(1 - sin²(x)) - 3sin(x) = 0

 

2 - 2sin²(x) - 3sin(x) = 0

 

-2sin²(x) - 3sin(x) + 2 = 0

 

 

We end up with a quadratic equation.

 

 

Let  q = sin(x)

 

-2q² - 3q + 2 = 0

 

 

Use the quadratic formula to solve for q.  Since we know the formula, lets go on ahead and plug in the values.

 

q = (3 ± √(9 - 4(-4))) / -4

 

q = (3 ± √(25)) / -4

 

q = (3 ± 5) / -4

 

q = 8 / -4 =  -2

 

q = -2 / -4  = 1/2

 

 

 sin(x) = -2                   and         sin(x) = 1/2

 

No solution here                           x = π/6

 

                                                  x = π - (π/6) = (6π - π) / 6 = 5π/6

 

 

Our solutions are

 

x = π/6

x = 5π/6