Respuesta :
Lets solve the trigonometric equation [tex]2sec^2(x)-tan^4(x)=1[/tex]
Step 1. Use the trigonometric identity: [tex]sec^2(x)=1+tan^2(x)[/tex]:
[tex]2[1+tan^2(x)]-tan^4(x)=1[/tex]
[tex]2+2tan^2(x)-tan^4(x)=1[/tex]
Step 2. Subtract 1 from both sides of the equation:
[tex]2-1+2tan^2(x)-tan^4(x)=1-1[/tex]
[tex]1+2tan^2(x)-tan^4(x)=0[/tex]
Step 3. Substitute [tex]tan^2(x)=u[/tex]:
[tex]1+2u-u^2=0[/tex]
Step 3. Multiply both sides of the equation by -1:
[tex]-1(1+2u-u^2)=0(-1)[/tex]
[tex]u^2-2-1=0[/tex]
Step 4. Solve for [tex]u[/tex]:
[tex]u=1+ \sqrt{2} ,u=1- \sqrt{2} [/tex]
Step 5. Substitute [tex]tan^2(x)=u[/tex]:
[tex]tan^2(x)=1+ \sqrt{2} ,tan^2(x)=1- \sqrt{2} [/tex]
Step 6. Solve for [tex]x[/tex]:
[tex] \sqrt{tan^2(x)} =(+/-) \sqrt{1+ \sqrt{2}} , \sqrt{tan^2(x)}= (+/-)\sqrt{1- \sqrt{2} } [/tex]
[tex]tan(x)=\sqrt{1+ \sqrt{2}},-\sqrt{1+ \sqrt{2}},\sqrt{1- \sqrt{2} },-\sqrt{1- \sqrt{2} }[/tex]
Since [tex]\sqrt{1- \sqrt{2} }[/tex] is not a real root, we can rule out [tex]\sqrt{1- \sqrt{2} },-\sqrt{1- \sqrt{2} }[/tex]
[tex]tan(x)=\sqrt{1+ \sqrt{2}},-\sqrt{1+ \sqrt{2}}[/tex]
[tex]x=arctan(\sqrt{1+ \sqrt{2}})+k \pi ,x=arctan(-\sqrt{1+ \sqrt{2}})+k \pi [/tex]
[tex]x=0.9989+k \pi ,x=-0.9989+k \pi [/tex]
We can conclude that the exact solution of the equation are [tex]x=arctan(\sqrt{1+ \sqrt{2}})+k \pi [/tex] and [tex]x=arctan(-\sqrt{1+ \sqrt{2}})+k \pi[/tex], which are approximately [tex]x=0.9989+k \pi [/tex] and [tex]x=-0.9989+k \pi [/tex] respectively.
Step 1. Use the trigonometric identity: [tex]sec^2(x)=1+tan^2(x)[/tex]:
[tex]2[1+tan^2(x)]-tan^4(x)=1[/tex]
[tex]2+2tan^2(x)-tan^4(x)=1[/tex]
Step 2. Subtract 1 from both sides of the equation:
[tex]2-1+2tan^2(x)-tan^4(x)=1-1[/tex]
[tex]1+2tan^2(x)-tan^4(x)=0[/tex]
Step 3. Substitute [tex]tan^2(x)=u[/tex]:
[tex]1+2u-u^2=0[/tex]
Step 3. Multiply both sides of the equation by -1:
[tex]-1(1+2u-u^2)=0(-1)[/tex]
[tex]u^2-2-1=0[/tex]
Step 4. Solve for [tex]u[/tex]:
[tex]u=1+ \sqrt{2} ,u=1- \sqrt{2} [/tex]
Step 5. Substitute [tex]tan^2(x)=u[/tex]:
[tex]tan^2(x)=1+ \sqrt{2} ,tan^2(x)=1- \sqrt{2} [/tex]
Step 6. Solve for [tex]x[/tex]:
[tex] \sqrt{tan^2(x)} =(+/-) \sqrt{1+ \sqrt{2}} , \sqrt{tan^2(x)}= (+/-)\sqrt{1- \sqrt{2} } [/tex]
[tex]tan(x)=\sqrt{1+ \sqrt{2}},-\sqrt{1+ \sqrt{2}},\sqrt{1- \sqrt{2} },-\sqrt{1- \sqrt{2} }[/tex]
Since [tex]\sqrt{1- \sqrt{2} }[/tex] is not a real root, we can rule out [tex]\sqrt{1- \sqrt{2} },-\sqrt{1- \sqrt{2} }[/tex]
[tex]tan(x)=\sqrt{1+ \sqrt{2}},-\sqrt{1+ \sqrt{2}}[/tex]
[tex]x=arctan(\sqrt{1+ \sqrt{2}})+k \pi ,x=arctan(-\sqrt{1+ \sqrt{2}})+k \pi [/tex]
[tex]x=0.9989+k \pi ,x=-0.9989+k \pi [/tex]
We can conclude that the exact solution of the equation are [tex]x=arctan(\sqrt{1+ \sqrt{2}})+k \pi [/tex] and [tex]x=arctan(-\sqrt{1+ \sqrt{2}})+k \pi[/tex], which are approximately [tex]x=0.9989+k \pi [/tex] and [tex]x=-0.9989+k \pi [/tex] respectively.
You can use the fact that the range of tangent function is whole set of real numbers.
The solutions to the given equation are
[tex]x = \pm \dfrac{\pi}{3 }+ n\pi ; \: n \in \mathbb Z\\[/tex]
What are Pythagorean identities ?
[tex]sin^2(\theta) + cos^2(\theta) = 1\\\\1 + tan^2(\theta) = sec^2(\theta)\\\\1 + cot^2(\theta) = csc^2(\theta)[/tex]
It is a fact that tangent ratio has range as all real numbers. We can use this fact along with the second Pythagorean identity to get to the solution of the given equation.
The given equation is [tex]2sec^2x-tan^4x=-1[/tex]
Using the second Pythagorean identity, we get the equation as
[tex]2\sec^2x-\tan^4x=-1\\\\2(1 + \tan^2x) - (\tan^2x)^2= -1\\\\(\tan^2x)^2 -2\tan^2x -3 = 0[/tex]
Assuming [tex]y = tan^2x[/tex], then we get [tex]y \geq 0[/tex]
The equation becomes
[tex](\tan^2x)^2 -2\tan^2x -3 = 0\\\\y^2 - 2y - 3 = 0\\y-3y + y - 3 = 0\\y(y - 3) + 1(y-3) = 0\\(y+1)(y-3) = 0\\y = -1, y = 3[/tex]
As we know that y is non-negative, so only valid solution is y = 3
Thus,
[tex]y = tan^2(x) = 3\\tan(x) = \pm \sqrt{3}\\x = \tan^{-1}(\pm \sqrt{3})[/tex]
Thus,
[tex]x = tan^{-1}(\sqrt{3}) = 60^\circ + n\pi ; \: n \in \mathbb Z\\\\x = tan^{-1}(-\sqrt{3}) = -60^\circ + n\pi ; \: n \in \mathbb Z[/tex]
Thus, the solutions to the given equation are:
[tex]x = \pm 60^\circ + n\pi ; \: n \in \mathbb Z\\[/tex]
Converting to radians,
[tex]x = \pm \dfrac{\pi}{3 }+ n\pi ; \: n \in \mathbb Z\\[/tex]
Thus,The solutions to the given equation are
[tex]x = \pm \dfrac{\pi}{3 }+ n\pi ; \: n \in \mathbb Z\\[/tex]
Learn more about trigonometric ratios here:
https://brainly.com/question/22599614
