Question

find all solution of the given equation in the interval [0,2pi) tan(x/2)=sinx

Answer 1

@ mae23

here,

we can write
tan(x/2) as sin(x/2)/cos(x/2)
then ,
$$\huge \dfrac{\sin(x/2)}{\cos(x/2}=sinx$$

now, we know that sinx/2 can be written as 2sin(x/2)cos(x/2)

than

$$\huge \dfrac{\sin(x/2)}{\cos(x/2}=2sin(x/2)cos(x/2)$$

now, $$\huge 1=2\cos^2(x/2)$$

finally,

$$\huge \cos(x/2)=\dfrac {1}{\sqrt{2}}$$

we know that $$\huge \cos(\pi/4)= 1/\sqrt{2}$$

so,$$\huge x/2= \pi/4 $$

hence,
$$\huge x= \pi/2$$

Understood it :)

Answer 2

@mae23

Answer 3

$$ \huge \color{red} {\ sinx = 2\ sin(x/2)\ cos(x/2)}$$

Answer 4

yes . thank you

Answer 5

$$\huge\frak \color{red} {WELCOME........}$$

Answer 6

@praxer is the interval equation pi/2,pi?

Answer 7

no it is not.

Answer 8

u gave the interval to be $$\huge \color{red} {{{[0,\pi}]}}$$

Answer 9

sorry [0,2pi]

Answer 10

it is in [0,2pi]

Answer 11

oh? I taught 2pi is not included

Answer 12

see $$\huge \pi/2$$ lies in the interval you denoted.