Question

cos3A + sin (2A - 7π/6) = -2
find the number of general solutions.


with explanation please (process)

class 11
trigonometry
maths

" don't beg for mark as brainliest, the correct answer and process will be marked as brainliest "​

Answer 1

SolutioN :

$\tt \dagger \: \: \: \: \: cos \: 3x + sin \: \bigg(2x - \dfrac{7 \pi}{6} \bigg) = - 2.$

• We know, The values of Sin θ and Cos θ range between [ - 1 , 1 ]
• And, When the value of Sin θ is - 1 and Cos θ is - 1 then, RHS shoud be satisfied.

$\tt : \implies Cos \: 3x = - 1.$

$\tt : \implies Cos \: 3x = Cos \:180.$

$\tt : \implies Cos \: 3x = \pi.$

$\tt : \implies Cos \: x = \dfrac{ \pi}{3}$

$\rule{90}2$

$\tt : \implies Sin \:\bigg(2x-\dfrac{7\pi}{6}\bigg) = - 1.$

• Now, We need to Putting the value of x in it. And let's prove RHS.

$\tt : \implies Sin \:\bigg(2x-\dfrac{7\pi}{6}\bigg)$

$\tt : \implies Sin \:\bigg(2 \times \dfrac{ \pi}{3} -\dfrac{7\pi}{6}\bigg)$

$\tt : \implies Sin \:\bigg( \dfrac{ 2\pi}{3} -\dfrac{7\pi}{6}\bigg)$

$\tt : \implies Sin \:\bigg(\dfrac{4 \pi - 7\pi}{6}\bigg)$

$\tt : \implies Sin \:\bigg(\dfrac{ - 3\pi}{6}\bigg)$

$\tt : \implies Sin \:\bigg(\dfrac{ - \cancel3\pi}{ \cancel6}\bigg)$

$\tt : \implies - Sin \:\bigg(\dfrac{\pi}{2}\bigg)$

$\tt : \implies - Sin \:90$

$\tt : \implies - 1.$

Hence, LHS = RHS.

Now, Let's Find there General Solution.

$\tt \dagger \: \: \: \: \: General \: Solution \: Of \: x.$

$\tt \longmapsto x = 2n \pi+ \dfrac{ \pi}{ 3}$

$\tt \longmapsto x = \dfrac{6n \pi + \pi}{ 3}$

$\tt \longmapsto x = \dfrac{ 6 \pi(n + 1)}{ 3}$

Therefore, the General Solution of given Equation is 6π ( n + 1 )/3.