Question

2) Find general solution of sec 0 = root 2
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Answer 1

Answer:

$\huge\boxed{\theta=\pm45^o+360^ok\to\theta=\pm\dfrac{\pi}{4}+2k\pi}\\\boxed{k\in\mathbb{Z}}$

Step-by-step explanation:

$\sec\theta=\dfrac{1}{\cos\theta}$

$\sec\theta=\sqrt2\\\\\dfrac{1}{\cos\theta}=\sqrt2\\\\\dfrac{1}{\cos\theta}=\dfrac{\sqrt2}{1}\qquad|\text{cross multiply}\\\\\sqrt2\cos\theta=1\qquad|\text{multiply both sides by}\ \sqrt2\\\\2\cos\theta=\sqrt2\qquad|\text{divide both sides by 2}\\\\\cos\theta=\dfrac{\sqrt2}{2}\to\theta=\pm45^o+360^ok\to\theta=\pm\dfrac{\pi}{4}+2k\pi$