What is the general solution for sin(x)+ cos(x)=-1?

Question

What is the general solution for sin(x)+ cos(x)=-1?

2 Answers

Impact of this question

9828 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-08T06:06:24

$x = n π - {(- 1)}^{n} (\frac{π}{4}) - \frac{π}{4}$ $x=npi-(-1)^n(pi/4)-pi/4$

Explanation:

As $sin x + cos x = - 1$ $sinx+cosx=-1$ , we have

$\sqrt{2} (sin x \cdot \frac{1}{\sqrt{2}} + cos x \cdot \frac{1}{\sqrt{2}}) = - 1$ $sqrt2(sinx*1/sqrt2+cosx*1/sqrt2)=-1$

or $sin x cos (\frac{π}{4}) + cos x sin (\frac{π}{4}) = - \frac{1}{\sqrt{2}}$ $sinxcos(pi/4)+cosxsin(pi/4)=-1/sqrt2$

or $sin (x + \frac{π}{4}) = sin (- \frac{π}{4})$ $sin(x+pi/4)=sin(-pi/4)$

Now general solution of $sin x = sin α$ $sinx=sinalpha$ is $x = n π \pm {(- 1)}^{n} α$ $x=npi+-(-1)^nalpha$

Hence, general solution is

$x + \frac{π}{4} = n π + {(- 1)}^{n} (- \frac{π}{4})$ $x+pi/4=npi+(-1)^n(-pi/4)$

or $x = n π + {(- 1)}^{n} (- \frac{π}{4}) - \frac{π}{4}$ $x=npi+(-1)^n(-pi/4)-pi/4$

or $x = n π - {(- 1)}^{n} (\frac{π}{4}) - \frac{π}{4}$ $x=npi-(-1)^n(pi/4)-pi/4$

Answer 2 · 2018-03-08T07:01:08

$x = \frac{3}{2} π + 2 π n or π + 2 π n$ $x=3/2 pi +2pin " or " pi + 2pin$

Explanation:

$sin (x) + cos (x) = - 1$ $sin(x)+cos(x)=-1$

$\Rightarrow sin (x) = - 1 - cos (x)$ $=>sin(x)=-1-cos(x)$ equation-1

We have an identity

${sin}^{2} (x) + {cos}^{2} (x) = 1$ $sin^2(x)+cos^2(x)=1$

Use this to find the value of $sin (x)$ $sin(x)$

$\Rightarrow {sin}^{2} (x) = 1 - {cos}^{2} (x)$ $=> sin^2(x)=1-cos^2(x)$

$\Rightarrow sin (x) = \pm \sqrt{1 - {cos}^{2} (x)}$ $=> sin(x)=+-sqrt(1-cos^2(x))$

We got two values for $sin (x)$ $sin(x)$

$+ \sqrt{1 - {cos}^{2} (x)} and - \sqrt{1 - {cos}^{2} (x)}$ $+sqrt(1-cos^2(x)) "and" -sqrt(1-cos^2(x))$

Put them one by one in equation-1.

$\Rightarrow + \sqrt{1 - {cos}^{2} (x)} = - 1 - cos (x)$ $=> +sqrt(1-cos^2(x))=-1-cos(x)$

Squaring both sides

$\Rightarrow 1 - {cos}^{2} (x) = 1 + {cos}^{2} (x) + 2 cos (x)$ $=> 1-cos^2(x)=1+cos^2(x)+2cos(x)$

$\Rightarrow 0 = 2 {cos}^{2} (x) + 2 cos (x)$ $=> 0=2cos^2(x)+2cos(x)$

Divide by two both sides

$\Rightarrow 0 = {cos}^{2} (x) + cos (x)$ $=> 0=cos^2(x)+cos(x)$

$\Rightarrow 0 = cos (x) (cos (x) + 1)$ $=> 0=cos(x)(cos(x)+1)$

It gives $cos (x) = 0$ $cos(x)=0$

We get $sin (x) = - 1$ $sin(x)=-1$

The solution for this is

$x = \frac{3}{2} π + 2 π n$ $x = 3/2 pi + 2pin$

Here , $π = 180^{\circ}$ $pi = 180^@$ and n is any integer.

Now , we also get $cos (x) + 1 = 0$ $cos(x)+1=0$

$\Rightarrow cos (x) = - 1$ $=> cos(x)=-1$

It gives $sin (x) = 0$ $sin(x)=0$ according to the given equation.

The solution for this is

$x = π + 2 π n$ $x= pi + 2pin$

If you put $sin (x) = - \sqrt{1 - {cos}^{2} (x)}$ $sin(x)=-sqrt(1-cos^2(x))$ you get the same results.

Hope it helps :)

Science

Math

Humanities

... and beyond

What is the general solution for sin(x)+ cos(x)=-1?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question