Use the identity cos2θ+sin2θ=1\cos^2\theta + \sin^2\theta = 1cos2θ+sin2θ=1 to prove that sec2θ=1+tan2θ\sec^2\theta = 1 + \tan^2\thetasec2θ=1+tan2θ
Solve, for 0≤θ≤3600 \leq \theta \leq 3600≤θ≤360, the equation,
3tan2θ+5secθ+1=0 3\tan^2\theta + 5\sec\theta + 1 = 0 3tan2θ+5secθ+1=0Give your answers to 1 decimal place.