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3 years ago

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

Mathematics

2 answers:

Kazeer [188]3 years ago

7 0

the answer is A on edg

dangina [55]3 years ago

4 0

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.

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More Calculus! (I'm so sorry)

Olenka [21]

Recall that converting from Cartesian to polar coordinates involves the identities

$\begin{cases}y(r,\phi)=r\sin\phi\\x(r,\phi)=r\cos\phi\end{cases}$

As a function in polar coordinates, $r$ depends on $\phi$ , so you can write $r=r(\phi)$ .

Differentiating the identities with respect to $\phi$ gives

$\begin{cases}\dfrac{\mathrm dy}{\mathrm d\phi}=\dfrac{\mathrm dr}{\mathrm d\phi}\sin\phi+r\cos\phi\\\\\dfrac{\mathrm dx}{\mathrm d\phi}=\dfrac{\mathrm dr}{\mathrm d\phi}\cos\phi-r\sin\phi\end{cases}$

The slope of the tangent line to $r(\phi)$ is given by

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm d\phi}}{\frac{\mathrm dx}{\mathrm d\phi}}=\dfrac{\frac{\mathrm dr}{\mathrm d\phi}\sin\phi+r\cos\phi}{\frac{\mathrm dr}{\mathrm d\phi}\cos\phi-r\sin\phi}$

Given $r(\phi)=3\cos\phi$ , you have $\dfrac{\mathrm dr}{\mathrm d\phi}=-3\sin\phi$ . So the tangent line to $r(\phi)$ has a slope of

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{-3\sin^2\phi+3\cos^2\phi}{-3\sin\phi\cos\phi-3\cos\phi\sin\phi}=\dfrac{3\cos2\phi}{-3\sin2\phi}=-\cot2\phi$

When $\phi=120^\circ=\dfrac{2\pi}3\text{ rad}$ , the tangent line has slope

$\dfrac{\mathrm dy}{\mathrm dx}=-\cot\dfrac{4\pi}3=-\dfrac1{\sqrt3}$

This line is tangent to the point $(r,\phi)=\left(-\dfrac32,\dfrac{2\pi}3\right)$ which in Cartesian coordinates is equivalent to $(x,y)=\left(\dfrac34,-\dfrac{3\sqrt3}4\right)$ , so the equation of the tangent line is

$y+\dfrac{3\sqrt3}4=-\dfrac1{\sqrt3}\left(x-\dfrac34\right)$

In polar coordinates, this line has equation

$r\sin\phi+\dfrac{3\sqrt3}4=-\dfrac1{\sqrt3}\left(r\cos\phi-\dfrac34\right)$
$\implies r=-\dfrac{3\sqrt3}{2\sqrt3\cos\phi+6\sin\phi}$

The tangent line passes through the y-axis when $x=0$ , so the y-intercept is $\left(0,-\dfrac{\sqrt3}2\right)$ .

The vector from this point to the point of tangency on $r(\phi)$ is given by the difference of the vector from the origin to the y-intercept (which I'll denote $\mathbf a$ ) and the vector from the origin to the point of tangency (denoted by $\mathbf b$ ). In the attached graphic, this corresponds to the green arrow.

$\mathbf b-\mathbf a=\left(\dfrac34,-\dfrac{3\sqrt3}4\right)-\left(0,-\dfrac{\sqrt3}2\right)=\left(\dfrac34,-\dfrac{\sqrt3}4\right)$

The angle between this vector and the vector pointing to the point of tangency is what you're looking for. This is given by

$\mathbf b\cdot(\mathbf b-\mathbf a)=\|\mathbf b\|\|\mathbf b-\mathbf a\|\cos\theta$
$\dfrac98=\dfrac{3\sqrt3}4\cos\theta$
$\implies\theta=\dfrac\pi6\text{ rad}=30^\circ$

The second problem is just a matter of computing the second derivative of $\phi$ with respect to $t$ and plugging in $t=2$ .

$\phi(t)=2t^3-6t$
$\phi'(t)=6t^2-6$
$\phi''(t)=12t$
$\implies\phi''(2)=24$