Question

2cos^2x + sinx-1=0

a. Explain why this equation cannot be factored.
b. Use a trigonometric identity to change the equation into one that can be factored.
c. Factor the equation.
d. Determine all solutions in the interval 0≤x≤2π.

Answer 1

@Luigi0210

Answer 2

@bagajr ??

Answer 3

can you send some things so I can start looking at it ?? @bagajr

Answer 4

@kc_kennylau

Answer 5

Firstly, be familiarized with this identity: \(\sin^2x+\cos^2x=1\)

Answer 6

Corollary: \(\sin^2x=1-\cos^2x\) and \(\cos^2x=1-\sin^2x\).

Answer 7

You may want to convert the question into an ordinary quadratic equation.

Answer 8

But why cant it be factored??

Answer 9

@bagajr

Answer 10

You will know why when you convert the question into an ordinary quadratic equation by changing \(\cos^2x\) to \(1-\sin^2x\).

Answer 11

a) Because cos^2x and sinx are not the same variable.
b) Use the following identity to solve for cos^2x\[\sin^{2}(x)+\cos^{2}(x)=1\]\[\cos^{2}(x) =1-\sin^{2}(x)\]Substitute cos^2x for this new expression.\[2\cos^{2}(x)+\sin(x)-1=0\]\[2[1-\sin^{2}(x)]+\sin(x)-1=0\]\[2-2\sin^{2}(x)+\sin(x)-1=0\]\[-2\sin^{2}(x)+\sin(x)+1=0\]\[2\sin^{2}(x)-\sin(x)-1=0\]c) This isn't necessary, but you can make factoring easier by setting y=sin(x)\[2y^{2}-y-1=0\]\[(2y+1)(y-1)=0\]\[(2\sin(x)+1)(\sin(x)-1)=0\]d) When you multiply two things to get 0, one of those things must be 0.\[2\sin(x)+1=0\]\[\sin(x)-1=0\]Solve both equations to get that sin(x) is -1/2 or 1. Use unit circle to find solutions, which are pi/2, 5pi/4, and 7pi/4.