Question

Number of solutions of the equation sinx = cos 4x for 0<=x<=π​

Answer 1

Answer:

sin

4

x+cos

4

x=sinx.cosx

⇒(sin

2

x+cos

2

x)−2sin

2

.cos

2

x=sinx.cosx

⇒1−

2

sin

2

2x

=

2

sin2x

⇒sin

2

2x+sin2x−2=0

⇒(sin2x−1)(sin2x+2)=0

=sin2x=−2(notpossible)

=sin2x=1

=2x=

2

π

,

2

5π

=x=

4

π

,

4

5π

I hope help you

Step-by-step explanation:

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