How do I solve this questions?

Question

How do I solve this questions?

Solve the given equation:
1) $tan θ - 3 cot θ = 0$ $tantheta-3cottheta=0$
2) $cos θ - sin θ = 1$ $costheta-sintheta=1$

2 Answers

Impact of this question

1357 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-10T11:03:33

For the equation $cos (θ) - sin (θ) = 1$ $cos(theta)-sin(theta)=1$ , the solution is $θ = 2 k π$ $theta=2kpi$ and $- \frac{π}{2} + 2 k π$ $-pi/2+2kpi$ for integers $k$ $k$

Explanation:

The second equation is $cos (θ) - sin (θ) = 1$ $cos(theta)-sin(theta)=1$ .

Consider the equation $sin (\frac{π}{4}) cos (θ) - cos (\frac{π}{4}) sin (θ) = \frac{\sqrt{2}}{2}$ $sin(pi/4)cos(theta)-cos(pi/4)sin(theta)=sqrt(2)/2$ . Notice that this is equivalent to the previous equation as $sin (\frac{π}{4}) = cos (\frac{π}{4}) = \frac{\sqrt{2}}{2}$ $sin(pi/4)=cos(pi/4)=sqrt(2)/2$ .

Then, using the fact that $sin (α \pm β) = sin (α) cos (β) \pm cos (α) sin (β)$ $sin(alphapmbeta)=sin(alpha)cos(beta)pmcos(alpha)sin(beta)$ , we have the equation:
$sin (\frac{π}{4} - θ) = \frac{\sqrt{2}}{2}$ $sin(pi/4-theta)=sqrt(2)/2$ .

Now, recall that $sin (x) = \frac{\sqrt{2}}{2}$ $sin(x)=sqrt(2)/2$ when $x = \frac{π}{4} + 2 k π$ $x=pi/4+2kpi$ and $x = \frac{3 π}{4} + 2 k π$ $x=(3pi)/4+2kpi$ for integers $k$ $k$ .

Thus,
$\frac{π}{4} - θ = \frac{π}{4} + 2 k π$ $pi/4-theta=pi/4+2kpi$
or
$\frac{π}{4} - θ = \frac{3 π}{4} + 2 k π$ $pi/4-theta=(3pi)/4+2kpi$

Finally, we have $θ = 2 k π$ $theta=2kpi$ and $- \frac{π}{2} + 2 k π$ $-pi/2+2kpi$ for integers $k$ $k$ .

Answer 2 · 2018-02-10T11:10:51

For the equation $tan (θ) - 3 cot (θ) = 0$ $tan(theta)-3cot(theta)=0$ , the solution is $θ = \frac{π}{3} + k π$ $theta=pi/3+kpi$ or $θ = \frac{2 π}{3} + k π$ $theta=(2pi)/3+kpi$ for integers $k$ $k$ .

Explanation:

Consider the first equation $tan (θ) - 3 cot (θ) = 0$ $tan(theta)-3cot(theta)=0$ . We know that $tan (θ) = \frac{1}{cot (θ)} = \frac{sin (θ)}{cos (θ)}$ $tan(theta)=1/cot(theta)=sin(theta)/cos(theta)$ .

Thus, $\frac{sin (θ)}{cos (θ)} - \frac{3 cos (θ)}{sin (θ)} = 0$ $sin(theta)/cos(theta)-(3cos(theta))/sin(theta)=0$ .

Then, $\frac{{sin}^{2} (θ) - 3 {cos}^{2} (θ)}{sin (θ) cos (θ)} = 0$ $(sin^2(theta)-3cos^2(theta))/(sin(theta)cos(theta))=0$ .

Now, if $sin (θ) cos (θ) \neq 0$ $sin(theta)cos(theta)≠0$ , we can safely multiply both sides by $sin (θ) cos (θ)$ $sin(theta)cos(theta)$ . This leaves the equation:
${sin}^{2} (θ) - 3 {cos}^{2} (θ) = 0$ $sin^2(theta)-3color(red)(cos^2(theta))=0$

Now, use the identity ${cos}^{2} (θ) = 1 - {sin}^{2} (θ)$ $cos^2(theta)=color(red)(1-sin^2(theta))$ into the red part of the equation above. Substituting this in gives us:
${sin}^{2} (θ) - 3 (1 - {sin}^{2} (θ)) = 0$ $sin^2(theta)-3(color(red)(1-sin^2(theta)))=0$
$4 {sin}^{2} (θ) - 3 = 0$ $4sin^2(theta)-3=0$
${sin}^{2} (θ) = \frac{3}{4}$ $sin^2(theta)=3/4$
$sin (θ) = \pm \frac{\sqrt{3}}{2}$ $sin(theta)=pmsqrt(3)/2$

The solution is thus $θ = \frac{π}{3} + k π$ $theta=pi/3+kpi$ or $θ = \frac{2 π}{3} + k π$ $theta=(2pi)/3+kpi$ for integers $k$ $k$ .

(Recall that we required $sin (θ) cos (θ) \neq 0$ $sin(theta)cos(theta)≠0$ . None of the solutions above would give us $sin (θ) cos (θ) = 0$ $sin(theta)cos(theta)=0$ , so we're fine here.)

Science

Math

Humanities

... and beyond

How do I solve this questions?

Solve the given equation:
1) $tan θ - 3 cot θ = 0$ $tantheta-3cottheta=0$
2) $cos θ - sin θ = 1$ $costheta-sintheta=1$

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do I solve this questions?

Solve the given equation: 1) tanθ−3cotθ=0tantheta-3cottheta=0 2) cosθ−sinθ=1costheta-sintheta=1

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

Solve the given equation:
1) $tan θ - 3 cot θ = 0$ $tantheta-3cottheta=0$
2) $cos θ - sin θ = 1$ $costheta-sintheta=1$