# Please help me with trig. Please explain how you got the answer so i can learn.?

Find all exact solutions that exist on the interval [0, 2π).

tan(x) sin(x) + sin(−x) = 0

### 3 Answers

- Pramod KumarLv 72 months ago
Answer is x = 0, π/4, π , 5π/4 , 2π

See how --

tan(x) sin(x) + sin(−x) = 0

We know that sin ( - x ) = - sin x

tan(x) sin(x) + sin(−x) = 0

=> tan(x) sin(x) - sin(x) = 0

=> ( sin x/cos x ) * ( sin x ) - sin x = 0

=> ( sin x ) ( sin x/cos x - 1 ) = 0

=> ( sin x ) ( tan x - 1 ) = 0

=> ( sin x ) = 0 Or tan x = 1

For sin x = 0 ...... x = 0 , π , 2π

For tan x = +1 ....... x = π/4 , 5π/4

Hence x = 0, π/4, π , 5π/4 , 2π .............. Answer

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- Pramod KumarLv 72 months agoReport
Plz indicate the mistakes.

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- rotchmLv 72 months ago
You (should) know that sin(-x) = -sin(x). Know this BY HEART.

Thus, your eqs becomes

tan(x)sin(x) - sin(x) = 0. Now, factor out sin(x) to get

sin(x) * (tan(x) - 1) = 0

Thus sin(x) = 0 and tan(x) - 1 = 0.

Can you solve each of these now?

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