Question

√3sinx=cosx then find 3cos²x+2cosx/3cosx+2​

Answer 1

Step-by-step explanation:

Simplification of the equation:

3cos2x−3sin2x−3–√2sinxcosx=0

3(cos2x−sin2x)−3–√(2sinxcosx)=0

[Trigonometric identities]

∵cos2x−sin2x=cos2x

2sinxcosx=sin2x

3(cos2x)−3–√(sin2x)=0

3(cos2x)=3–√sin2x

33–√=sin2xcos2x

3–√=tan2x

tan−1(3–√)=2x

(According to the property of inverse trigonometric functions)

tan−1(tanπ3)=2x

π3=2x

π6=x

It is the solution of this problem