What are the solution in the interval [0, 2pie]? sec^2x-tan^2x=0

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Answer 1 · 2018-04-04T07:41:40

See explanation

We want to solve the equation

${sec}^{2} (x) - {tan}^{2} (x) = 0$ $sec^2(x)-tan^2(x)=0$

To get a better idea, write this in terms of sine and cosine

$\frac{1}{{cos}^{2} (x)} - \frac{{sin}^{2} (x)}{{cos}^{2} (x)} = 0$ $1/cos^2(x)-sin^2(x)/cos^2(x)=0$

To simplify multiply both sides by ${cos}^{2} (x)$ $cos^2(x)$

$1 - {sin}^{2} (x) = 0$ $1-sin^2(x)=0$

Thus

${sin}^{2} (x) = 1$ $sin^2(x)=1$

$\Rightarrow sin (x) = \pm \sqrt{1}$ $=>sin(x)=+-sqrt(1)$

$\Rightarrow sin (x) = \pm 1$ $=>sin(x)=+-1$

Now i will leave it for you to find the solutions