Solve: (1-√3tanθ)+1+√3 = √3sec^2θ Where (0≤θ≤360)
recall that sec^2 Ø = tan^2 Ø + 1
(1-√3tanθ)+1+√3 = √3sec^2θ
1 - √3tanØ + 1 + √3 = √3(tan^2 Ø + 1)
- √3tanØ + √3 = √3tan^2 Ø + √3
tan^2 Ø + tanØ = 0
tanØ(tanØ + 1) = 0
tanØ = 0 or tanØ = -1
Ø = 0 or Ø = 135° or Ø = 315°
in the solution of tanØ = 0
I should have included Ø = 180°
so Ø = 0°, 135°, 180°, 315°
To solve the equation: (1-√3tanθ)+1+√3 = √3sec^2θ, follow these steps:
Step 1: Simplify the left side of the equation:
(1-√3tanθ)+1+√3 = √3sec^2θ
Subtract 1 and √3 from both sides:
-√3tanθ = √3sec^2θ - 2
Step 2: Use trigonometric identities to simplify the equation.
Recall the trigonometric identity: sec^2θ = 1 + tan^2θ
Replace sec^2θ in the equation:
-√3tanθ = √3(1 + tan^2θ) - 2
Step 3: Distribute √3 on the right side of the equation:
-√3tanθ = √3 + 3tan^2θ - 2
Step 4: Rearrange the equation to gather all the terms on one side:
3tan^2θ + √3tanθ - √3 - 2 = 0
Step 5: Use the quadratic formula to solve for tanθ:
tanθ = (-b ± √(b^2 - 4ac))/2a
In this case, a = 3, b = √3, c = -2.
Plug these values into the quadratic formula and simplify:
tanθ = (-√3 ± √(3 + 24))/6
tanθ = (-√3 ± √27)/6
tanθ = (-√3 ± 3√3)/6
tanθ = (√3(-1 ± √3))/6
tanθ = (-1 ± √3)/2
Step 6: Find the values of θ that correspond to the tangent values.
Using the unit circle or a reference triangle, we can find the values of θ.
For tanθ = (-1 + √3)/2:
θ = 150 degrees
For tanθ = (-1 - √3)/2:
θ = 330 degrees
Therefore, the solutions for θ are 150 degrees and 330 degrees.