Question

sinx=3/4 then find general solution of cotx ?

Answer 1

Answer:

x =  n π + cot⁻¹ ( √ 7 / 3 )

Step-by-step explanation:

Given :

sin x = 3  /4

= > cos x = √ ( 1 - sin² x )

= > cos x = √ ( 1 - 9 / 16 )

= > cos x = √ 7 / 4

Now cot x = cos x / sin x

cot x = ( √ 7 / 4 ) / ( 3 / 4 )

= > cot x = √ 7 / 3

cot x = 1 / tan x = √ 7 / 3

= > tan x = 3 / √ 7

Let say :

tan α = 3 / √ 7

= > α = tan⁻¹ ( 3 / √ 7 )

We know if :

tan Ф = tan α

= > Ф = n π + α

= > x = n π + tan⁻¹ ( 3 / √ 7 )

In term of cot as we know :

tan⁻¹ x = cot⁻¹ ( 1 / x )

= > x =  n π + cot⁻¹ ( √ 7 / 3 )

Hence we get required answer.

Answer 2

‌‍,

.

$x = n\pi + {cot}^{ - 1} ( \sqrt{7} \div 3 \\ \\ \\ sin \: x = 3 \div 4 \\ \\ = cos = x \sqrt{} (1 - {sin}^{2} x) \\ \\ = cosx = \sqrt{} (1 - 9 \div 16 ) \\ \\ = cos = \sqrt{} 7 \div 4 \\ \\ \\ cot \: x = cos \: x \div \: sin \: x \\ \\ cot \: x = ( \sqrt{} 7 \div 4) \div (3 \div 4) \\ \\ = cot \: x = \sqrt{} 7 \div 3 \\ \\ \\ cot \: x = 1 \div tan \: x = \sqrt{} 7 \div 3 \\ \tan \: x = 3 \div \sqrt{} 7 \\ \\ let \: \: see \\ \\ tan \: \: a \: = 3 \div \sqrt{} 7 \\ \\ = a \: = \: {tan}^{ - 1} (3 \div \sqrt{} 7) \\ \\ wee \: \: know \: \: if \\ \\ tan \: 0 = 3 = tan \: \: a \\ = 0 = n\pi + a \\ \\ = x = 2\pi \: \: {tan}^{ - 1} (3 \div \sqrt{} 7) \\ \\ \: in \: term \: of \: cot \: as \: we \: know \: \\ \\ \\ {tan}^{ - 1} x = {cot}^{ - 1} (1 \div x) \\ = x = n\pi + \: {cot}^{ - 1} ( \sqrt{} 7 \div 3)$