# How do I solve this trigonometric equation?

Solve:

2 sin^2a +sin a cos a(cot a - tan a)

### 8 Answers

- lenpol7Lv 71 month ago
CotA - Tan A = CosA / SinA - SinA/CosA

Subtracting

Cos^2A - Sin^2A /(CosASinA)

Re inserting

2Sin^2A + SinACosA( Cos^2A - Sin^2A) / (SinACosA)

Cancel down 'SinACosA

2Sin^2A + ( Cos^A - Sin^2A)

Use the Trig. Identity Cos^2A = 1 - Sin^2A

Substituting

2Sin^2A + ( 1 - Sin^2A - Sin^2A)

2Sin^2A + 1 - 2Sin^2A = '1' NB ( 2Sin^2A add to 'zero'.

- PinkgreenLv 71 month ago
Mistake: what you posted was not an equation, after

making correction & re-post it.

- husoskiLv 71 month ago
Step 1 is to realize that it's not an equation. It's an expression. There's nothing to solve for. The value of a can be anythign that makes both (tan a) and (cot a) defined.

You can simplify, though. If I don't see an identity I can use right away, I find it helps to write everything in terms of sines and cosines:

2 sin² a + (sin a)(cos a)(cot a - tan a) = 2 sin² a + (sin a)(cos a)[(cos a)/(sin a) - (sin a)/(cos a)]

= 2 sin² a + (cos² a - sin² a)

That last step multiplied (sin a)(cos a) into each term in [] brackets and canceled denominators. Finish with:

= sin² a + cos² a

= 1

That's only true when both cos a and sin a are nonzero, making both the original expression and the cancellation of (sin a)/(sin a) and (cos a)/(cos a) valid.

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- fcas80Lv 71 month ago
It's not an equation.

However, if you set it to zero, i think there are no solutions.

- nyphdinmdLv 71 month ago
work out the second term

sin(a)cos(a)(cot(a) - tan(a)) = sin(a)cos(a)( cos(a)/sin(a) - sin(a)/cos(a))

= cos^2(a) - sin^2(a)

then

2 sin^2a +sin a cos a(cot a - tan a) = 2*sin^2(a) + cos^2(a) - sin^2(a) = sin^2(a) +cos^2(a) = 1

- rotchmLv 71 month ago
You can't solve it because there's nothing to solve. To "solve" an equation you must have an equals sign.

If you meant to find the roots then here just factor out sin(a).

What does this imply?

To your update. Still your question doesn't make sense. Calculate what? What you type is like 5 + x, then you say calculate 5+x... well, its 5+x.

Type your question exactly as is it is posed to you. Provide a pic if need be.