θ \theta

Geometry Level 2

tan ⁡ 2 θ + sec ⁡ θ + 1 = 0 \large \tan^2 \theta + \sec \theta + 1 = 0

Solve the equation above for 0 < θ < 2 π 0<\theta <2\pi .

5 5 2 π 2\pi π 6 \dfrac{\pi}{6} π \pi

Nazmus Sakib by Nazmus sakib , 24, Bangladesh

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1 solution

Chew-Seong Cheong
Sep 26, 2017

tan ⁡ 2 θ + sec ⁡ θ + 1 = 0 sec ⁡ 2 θ + sec ⁡ θ = 0 sec ⁡ θ ( sec ⁡ θ + 1 ) = 0 sec ⁡ θ = − 1 Note that sec ⁡ θ = 0 has no real solution. θ = π for 0 < θ < 2 π \begin{aligned} {\color{#3D99F6}\tan^2 \theta} + \sec \theta + \color{#3D99F6} 1 & = 0 \\ {\color{#3D99F6}\sec^2 \theta} + \sec \theta & = 0 \\ \sec \theta ( \sec \theta + 1) & = 0 \\ \sec \theta & = - 1 & \small \color{#3D99F6} \text{Note that }\sec \theta = 0 \text{ has no real solution.} \\ \theta & = \boxed{\pi} & \small \color{#3D99F6} \text{for }0 < \theta < 2 \pi \end{aligned}

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