Question

what is the solution to 2cos^2 x+3sinx-3=0?

Answer 1

\(\bf 2{\color{blue}{ cos^2(x)}}+3sin(x)-3=0\implies 2[{\color{blue}{ 1-sin^2(x)}}]+3sin(x)-3=0
\\ \quad \\
2-2sin^2(x)+3sin(x)-3=0\implies 0=2sin^2(x)-3sin(x)+1\)

notice it's just a quadratic, thus solve it likewise

Answer 2

i mean for the values on the graph between 0 and 360

Answer 3

You still need to solve the above equation for x. Then, you list all of the x values that lie within the interval [0, 360 degrees).

Hint: try solving 2y^2 - 3y + 1 for y. Then apply the same technique to solving the equation in sin x for sin x, first, and for x, second.

Answer 4

Can you solve 2y^2 - 3y + 1 = 0? Please give this a try.

Answer 5

You could:
1. factor, or
2. use the Quadratic Formula, or
3. solve for y by graphing

Answer 6

2y^2 - 3y + 1 = 0.
Remind you of 2 Tips in quick solving quadratic equations
Tip 1. When a + b + c = 0, one real root is (1), and the other is (c/a)
Tip 2. When a - b + c = 0, one real root is (-1), and the other is (-c/a)
In the above example (Tip 1), one real root is y1 = 1 and y2 = c/a = 1/2
y1 = cos x = cos 0 and cos 2Pi --> x = .... and .....
y2 = cos x = 1/2 = cos Pi/3 and cos (2Pi - Pi/3) = 5Pi/3-> x = ....., and .....