A research team is modeling the vertical displacement, HHH, of a specialized underwater sensor. The displacement is given by the function
H(θ)=sin2θsecθ+cos2θcscθ+sinθ H(\theta) = \sin 2\theta \sec \theta + \cos 2\theta \csc \theta + \sin \theta H(θ)=sin2θsecθ+cos2θcscθ+sinθwhere θ\thetaθ is the tilt angle of the sensor.
Show that the expression for H(θ)H(\theta)H(θ) can be written as
H(θ)=sinθ+cscθ H(\theta) = \sin \theta + \csc \theta H(θ)=sinθ+cscθwhere sinθ≠0\sin \theta \neq 0sinθ=0 and cosθ≠0\cos \theta \neq 0cosθ=0.
A technician attempts to find the tilt angles where the displacement is exactly 4.254.254.25 units by solving the equation
sin2θsecθ+cos2θcscθ+sinθ=4.25 \sin 2\theta \sec \theta + \cos 2\theta \csc \theta + \sin \theta = 4.25 sin2θsecθ+cos2θcscθ+sinθ=4.25for 0∘≤θ≤360∘0^{\circ} \leq \theta \leq 360^{\circ}0∘≤θ≤360∘. They produce the following solution:
Step 1: sinθ+cscθ=4.25\sin \theta + \csc \theta = 4.25sinθ+cscθ=4.25
Step 2: sinθ+1sinθ=174\sin \theta + \frac{1}{\sin \theta} = \frac{17}{4}sinθ+sinθ1=417
Step 3: 4sin2θ−17sinθ+4=04\sin^{2} \theta - 17\sin \theta + 4 = 04sin2θ−17sinθ+4=0
Step 4: sinθ=4\sin \theta = 4sinθ=4 or sinθ=0.25\sin \theta = 0.25sinθ=0.25
Step 5: θ=14.5∘,165.5∘\theta = 14.5^{\circ}, 165.5^{\circ}θ=14.5∘,165.5∘
Explain why the technician should reject the value sinθ=4\sin \theta = 4sinθ=4 in Step 4.
Determine if there are any other reasons, based on the original expression's domain, why solutions might need to be rejected, and state the final correct solutions for the technician's equation in the range 0∘≤θ≤360∘0^{\circ} \leq \theta \leq 360^{\circ}0∘≤θ≤360∘.
