Question

Find the general solution of the equation- $\sec^22\mathrm x=1-\tan2\mathrm x$

Answer 1

Hint: The general solution of $\mathrm{tan\mathrm\theta}=\mathrm{tan\mathrm\alpha}$ is given by-
$\mathrm\theta=\mathrm{n\mathrm\pi}+\mathrm\alpha\;\left(\mathrm n=0,\;\pm1,\;\pm2,\;...\right).....\left(1\right)$
Also, the trigonometric properties required are-
$\sec^2\mathrm y-\tan^2\mathrm y=1$

Complete step-by-step answer:
We have to solve the equation for x. First we will convert the whole equation in terms of tan2x-
$\sec^22\mathrm x=1-\tan2\mathrm x\\\\\mathrm{Using}\;\sec^2\mathrm y-\tan^2\mathrm y=1,\\\\\sec^2\mathrm y=1+\tan^2\mathrm y....\left(2\right)$
Using equation (2), we can write that-
$1+\tan^22\mathrm x=1-\tan2\mathrm x\\\\\tan^22\mathrm x+\tan2\mathrm x=0\\\\\tan2\mathrm x\left(\tan2\mathrm x+1\right)=0\\\\\tan2\mathrm x=0,\;\tan2\mathrm x=-1\\\\\tan2\mathrm x=0=\tan0\\\\\mathrm{Using}\;\mathrm{equation}\;\left(1\right),\\\2\mathrm x=\mathrm{n\mathrm\pi}+0\\\\\mathrm x=\dfrac{\mathrm{n\mathrm\pi}}2$
Also, we know that tan is negative between 90o and 180o, tan(90+45)=tan135o = -1
$\tan2\mathrm x=-1=\tan\dfrac{3\mathrm\pi}4\\\2\mathrm x=\mathrm{n\mathrm\pi}+\dfrac{3\mathrm\pi}4\\\\\mathrm x=\left(4\mathrm n+3\right)\dfrac{\mathrm\pi}8$
So, the general solution is-
$\mathrm x=\dfrac{\mathrm{n\mathrm\pi}}2,\;\left(4\mathrm n+3\right)\dfrac{\mathrm\pi}8,\;\mathrm{where}\;\mathrm n=0,\;\pm1,\;\pm2,\;...$

Note: To solve trigonometric equations, try and convert all the terms in one type of function. Also, try to remember the formula for general solution of all trigonometric functions. Even after getting the solution, substitute the value and check if the solution exists, because there might be an extra root forming while converting the trigonometric functions.