Question

BRAINLIESTTT ASAP! PLEASE HELP ME :)

Answer 1

Answer:

θ = 0.896 and 2.246.

Step-by-step explanation:

We will use the identity

sin²θ + cos²θ = 1

and the double angle formula

cos2θ = cos²θ - sin²θ = (1 - sin²θ) - sin²θ = 1 - 2sin²θ

1. Convert the equation to a more solvable form.

$\begin{array}{rcll}\sin\theta + 1 & = & \cos2\theta &\\\sin\theta + 1& = & 1-2\sin^{2}\theta & \text{Used the double angle formula}\\f(\theta) = 2\sin^{2}\theta + \sin\theta -2 & = & 0 & \text{Moved terms to left-hand side}\\2u^2 + u - 2 & = & 0 & \text{Substituted u for }\sin\theta\\\end{array}$

2. Solve the quadratic

a = 2; b = 1; c = -2

(a) Set up the equation.

$u =\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\=\dfrac{-1\pm\sqrt{(1^2 - 4\times 2 \times (-2)}}{2\times2}\\\\=\dfrac{-1\pm\sqrt{1 + 16}}{4}\\\\=\dfrac{-1\pm\sqrt{17}}{4}$

(b) Evaluate the roots

$\begin{array}{rcl}u = \dfrac{-1 - 4.123}{4} & \qquad & u = \dfrac{-1 + 4.123}{4}\\\\u = -\dfrac{5.123}{4} & \qquad & u =\dfrac{3.123}{4}\\\\u = -1.281& \qquad & u =0.7808\\\end{array}$

3. Solve for θ

(a) The negative root

u = sinθ = -1.281

sinθ cannot be less than -1. No solution.

(b) The positive root

If sin(x) = a, then

x = arcsin(x) + 2πn and

x = π - arcsin(a) + 2πn

(i) Case 1

u = sinθ = 0.7808

θ = arcsin(0.7808) + 2πn

θ = 0.8959 ± 2πn

If n = 0,  

θ = 0.8959

(ii) Case 2

θ = π - 0.8959 + 2πn

If n = 0,

θ = π - 0.8959 = 2.246

The solutions are θ = 0.8959 and θ = 2.246

The diagram below shows the graph of f(θ) with its roots at 0.896 and 2.246.