Question

Which shows all the exact solutions of 2sec^2x-tan^4x=-1 ? Give your answer in radians.

Answer 1

the answer is A on edg

Answer 2

Lets solve the trigonometric equation $2sec^2(x)-tan^4(x)=1$

Step 1. Use the trigonometric identity: $sec^2(x)=1+tan^2(x)$:
$2[1+tan^2(x)]-tan^4(x)=1$
$2+2tan^2(x)-tan^4(x)=1$

Step 2. Subtract 1 from both sides of the equation:
$2-1+2tan^2(x)-tan^4(x)=1-1$
$1+2tan^2(x)-tan^4(x)=0$

Step 3. Substitute $tan^2(x)=u$:
$1+2u-u^2=0$

Step 3. Multiply both sides of the equation by -1:
$-1(1+2u-u^2)=0(-1)$
$u^2-2-1=0$

Step 4. Solve for $u$:
$u=1+ \sqrt{2} ,u=1- \sqrt{2}$

Step 5. Substitute $tan^2(x)=u$:
$tan^2(x)=1+ \sqrt{2} ,tan^2(x)=1- \sqrt{2}$

Step 6. Solve for $x$:
$\sqrt{tan^2(x)} =(+/-) \sqrt{1+ \sqrt{2}} , \sqrt{tan^2(x)}= (+/-)\sqrt{1- \sqrt{2} }$
$tan(x)=\sqrt{1+ \sqrt{2}},-\sqrt{1+ \sqrt{2}},\sqrt{1- \sqrt{2} },-\sqrt{1- \sqrt{2} }$
Since $\sqrt{1- \sqrt{2} }$ is not a real root, we can rule out $\sqrt{1- \sqrt{2} },-\sqrt{1- \sqrt{2} }$
$tan(x)=\sqrt{1+ \sqrt{2}},-\sqrt{1+ \sqrt{2}}$
$x=arctan(\sqrt{1+ \sqrt{2}})+k \pi ,x=arctan(-\sqrt{1+ \sqrt{2}})+k \pi$
$x=0.9989+k \pi ,x=-0.9989+k \pi$

We can conclude that the exact solution of the equation are $x=arctan(\sqrt{1+ \sqrt{2}})+k \pi$ and $x=arctan(-\sqrt{1+ \sqrt{2}})+k \pi$, which are approximately $x=0.9989+k \pi$ and $x=-0.9989+k \pi$ respectively.