Question Find all angles, 0≤θ<360, that satisfy the equation below, to the nearest 10th of a degree. 4tan^2 θ+tanθ=−4tanθ−1

Answer: 135 and 315degrees Step-by-step explanation: Given the expression 4tan²θ+tanθ=−4tanθ−1 Equate to zero 4tan²θ+tanθ+4tanθ+1 = 0 4tan²θ+5tanθ+1 = 0 Let x = tanθ 4x²+5x+1 = 0 Factorize; 4x²+4x+x+1 = 0 4x(x+1)+1(x+1) = 0 4x+1 = 0 and x+1 = 0 4x = -1 and x = -1 x = -1/4 and -1 Since x = tanθ -1 =tanθ θ = arctan(-1) θ = -45degrees Since tan is negative in the 2nd and fourth quadrant; In the second quadrant θ = 180 – 45 = 135degrees In the fourth quadrant; θ = 360 – 45 = 315degrees Hence the required angles are 135 and 315degrees Reply

Answer:135 and 315degreesStep-by-step explanation:Given the expression4tan²θ+tanθ=−4tanθ−1

Equate to zero4tan²θ+tanθ+4tanθ+1 = 0

4tan²θ+5tanθ+1 = 0

Let x = tanθ

4x²+5x+1 = 0

Factorize;4x²+4x+x+1 = 0

4x(x+1)+1(x+1) = 0

4x+1 = 0 and x+1 = 0

4x = -1 and x = -1

x = -1/4 and -1

Since x = tanθ

-1 =tanθ

θ = arctan(-1)

θ = -45degrees

Since tan is negative in the 2nd and fourth quadrant;

In the second quadrantθ = 180 – 45 = 135degrees

In the fourth quadrant;θ = 360 – 45 = 315degrees

Hence the required angles are 135 and 315degrees