Solve the following equation for 0° ≤ θ < 360°. use the "^" key on the keyboard to indicate an exponent. for example, sin2x would be typed as sin^2x. be sure to show all your work. 2sin2x - cos2 x - 2 = 0

This is the guy aboves work, I just wrote it out so you can copy and paste easier.

2sin^2x-cos^2x-2=0

sin^2x+cos^2x=1

cos^2x=1-sin^2x

2sin^2x-(1-sin^2x)-2=0

2sin^2x -1 +sin^2x-2=0

3sin^2x-3=0

3sin^2x=3

sin^2x=1

sin x =(sqr root)1

sin x= + or - 1

sin x =1    sin x = -1

x= sin^-1(1)   x=sin^-1 (-1)

x=90+n360    x=270 + 360n

x=90, 270 (wrote down so it stands out more)

black lol

step-by-step explanation:

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i can tell you jamal now has 3 cookies xd

tommy has 2.

this made me lol.