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 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