Solve, for 0 ≤ θ < 360, the equation 2 tan^2 θ + sec θ = 1 giving your answer to 1 decimal place please help
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Hello, Originally Posted by sharp357 Solve, for 0 ≤ θ < 360, the equation 2 tan^2 θ + sec θ = 1 giving your answer to 1 decimal place please help Remember : $\displaystyle 1+\tan^2(x)=\sec^2(x)$ so your equation is : $\displaystyle 2\sec^2 \theta+\sec \theta-3=0$ this is a quadratic
ok thaks, i can do this now
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