How do you find the exact value of $cos x sin x + sin x = 0$ $cosxsinx+sinx=0$ in the interval $0 \leq x < 360$ $0<=x<360$ ?

Question

How do you find the exact value of $cos x sin x + sin x = 0$ $cosxsinx+sinx=0$ in the interval $0 \leq x < 360$ $0<=x<360$ ?

1 Answer

Impact of this question

10375 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-20T02:01:01

${0^{\circ}, 180^{\circ}}$ ${0^@, 180^@}$

Explanation:

We can factor out a sine.

$sin x (cos x + 1) = 0$ $sinx(cosx + 1) = 0$

We now have two possible cases.

Case 1: $sin x = 0$ $sinx = 0$

$x = 0^{\circ}, 180^{\circ}$ $x = 0^@, 180^@$

Case 2: $cos x + 1 = 0$ $cosx + 1 = 0$

$cos x = - 1$ $cosx = -1$

$x = 180^{\circ}$ $x = 180^@$

$:.$ Our solution set in $0 <= x < 360$ is ${0^@, 180^@}$ .

Hopefully this helps!

Science

Math

Humanities

... and beyond

How do you find the exact value of $cos x sin x + sin x = 0$ $cosxsinx+sinx=0$ in the interval $0 \leq x < 360$ $0<=x<360$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you find the exact value of cosxsinx+sinx=0cosxsinx+sinx=0cosxsinx+sinx=0 in the interval 0<=x<3600≤x<3600<=x<360?

1 Answer

Explanation:

Related questions

Impact of this question

How do you find the exact value of $cos x sin x + sin x = 0$ $cosxsinx+sinx=0$ in the interval $0 \leq x < 360$ $0<=x<360$ ?