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 - cos2x - 2 = 0

2sin^2x+sin2x+cos2x=0 (1).

By using the trigonometric identities below :

sin2x=2sinxcosx

cos2x=cos^2x-sin^2x

We substitute the trigonometric identities into (1).

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

By combining like terms .

sin^2x+2sinxcosx+cos^2x=0(2)

The equation (2) is equivalent to the following expression (3).

(sinx+cosx)(sinx+cosx)=0 (3).

sinx+cosx=0

cosx=-sinx

divide both sides by cosx

1=-sinx/cosx

-1=sinx/cosx

sinx/cosx=tanx

substitute

-1=tanx

tanx=-1

tangent is negative in 2nd and 4th quadrants

tan135º=-1 (one answer)

tan315º=-1 (second answer)

Step-by-step explanation:

Please refer to the trigonometric identities used and explained above .

what are the options? ?

step-by-step explanation:

guess at question:

a vertical line intersects the graph of g(x,y) = 0 at two points. what conclusion follows about the function f(x) implicitly defined by g(x,f(x)) = 0?

the domain of f(x) cannot include the x coordinate of any such vertical line.

informally, f(x) is not a function, and

g(x,y)=0 cannot be solved for y uniquely.

the relation r = {(x,y) such that g(x,y) = 0} is a function if

for all (a,b) ∊ r and (c,d) ∊ r, a=c implies b=d.

step-by-step explanation: