If $12 \sin ^2 \alpha-4 \sin \alpha-1=0$ over the interval $0^{\circ} \leq \alpha \leq 180^{\circ}$, then $\alpha=$ ?
Answer (1)
Substitute x = sin α and rewrite the equation as a quadratic equation: 12 x 2 − 4 x − 1 = 0 .
Solve the quadratic equation using the quadratic formula, obtaining x = 2 1 and x = − 6 1 .
Find the angles α such that sin α = 2 1 within the interval 0 ∘ ≤ α ≤ 18 0 ∘ , which gives α = 3 0 ∘ and α = 15 0 ∘ .
Since sin α cannot be negative in the interval 0 ∘ ≤ α ≤ 18 0 ∘ , discard the solution sin α = − 6 1 . The final answer is 3 0 ∘ , 15 0 ∘ .
Explanation
Understanding the Problem We are given the equation 12 × \tensors sin 2 \tensors α − 4 × \tensors sin \tensors α − 1 = 0 and asked to find the values of α in the interval 0 ∘ ≤ α ≤ 18 0 ∘ that satisfy this equation. This looks like a quadratic equation in terms of sin α , so we can use the quadratic formula to solve for sin α .
Transforming the Equation Let x = sin α . Then the equation becomes 12 x 2 − 4 x − 1 = 0 . We can solve this quadratic equation for x using the quadratic formula: x = 2 a − b ± b 2 − 4 a c where a = 12 , b = − 4 , and c = − 1 .
Applying the Quadratic Formula First, let's calculate the discriminant: Δ = b 2 − 4 a c = ( − 4 ) 2 − 4 ( 12 ) ( − 1 ) = 16 + 48 = 64 Now, we can apply the quadratic formula: x = 24 4 ± 64 = 24 4 ± 8
Finding Possible Values for Sine Alpha We have two possible values for x : x 1 = 24 4 + 8 = 24 12 = 2 1 x 2 = 24 4 − 8 = 24 − 4 = − 6 1 Since x = sin α , we have sin α = 2 1 and sin α = − 6 1 .
Solving for Alpha when sin(alpha) = 1/2 For sin α = 2 1 , we need to find the values of α in the interval 0 ∘ ≤ α ≤ 18 0 ∘ . We know that α = arcsin ( 2 1 ) = 3 0 ∘ . Since sin ( 18 0 ∘ − x ) = sin x , another solution is 18 0 ∘ − 3 0 ∘ = 15 0 ∘ .
Solving for Alpha when sin(alpha) = -1/6 For sin α = − 6 1 , we need to find the values of α in the interval 0 ∘ ≤ α ≤ 18 0 ∘ . However, since the sine function is negative in the third and fourth quadrants, and we are only considering 0 ∘ ≤ α ≤ 18 0 ∘ , there are no solutions in this interval because sin α is non-negative in the first and second quadrants.
Final Answer Therefore, the solutions are α = 3 0 ∘ and α = 15 0 ∘ .
Examples
Understanding trigonometric equations is crucial in various fields such as physics and engineering. For example, when analyzing the motion of a pendulum, the angle it makes with the vertical direction can be modeled using trigonometric functions. Solving equations like the one above helps determine specific angles at certain points in time, which is essential for predicting the pendulum's behavior. Similarly, in electrical engineering, alternating current (AC) circuits involve sinusoidal functions, and solving trigonometric equations is necessary to analyze voltage and current values at different phases.
