Question

solve for x :- 4sinxcosx+2sinx+2cosx+1=0

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

The value of $x = n \pi \pm ( - 1 ) ^ { n + 1 } \frac { \pi } { 6 } \text { and } x = \frac { 2 \pi } { 3 } \pm 2 n \pi$ where n terms to be the integer.

Solution:

Given,  

$\begin{array} { c } { 4 \sin x \cos x + 2 \sin x + 2 \cos x + 1 = 0 } \\\\ { 2 \sin x ( 2 \cos x + 1 ) + 1 ( 2 \cos x + 1 ) = 0 } \\\\ { ( 2 \sin x + 1 ) ( 2 \cos x + 1 ) = 0 } \end{array}$

Either the value of x will be,

$\begin{array} { c } { ( 2 \sin x + 1 ) = 0 } \\\\ { \sin x = - \frac { 1 } { 2 } } \\\\ { x = n \pi \pm ( - 1 ) ^ { n + 1 } \frac { \pi } { 6 } } \end{array}$

We know that, if $\sin \theta = \frac { 1 } { 2 }$, this is satisfied by $\theta = \frac { \pi } { 6 } , \frac { 5 \pi } { 6 }$ etc. i.e., multiple of $\frac { \pi } { 6 } \text { is } n \pi \pm \frac { \pi } { 6 }$

For infinite solutions, this value will be wholly represented as $\theta = n \pi \pm ( - 1 ) ^ { n + 1 } \frac { \pi } { 6 }$, where n termed to be the integer.

Or the value of x will be,

$\begin{array} { c } { ( 2 \cos x + 1 ) = 0 } \\\\ { \cos x = - \frac { 1 } { 2 } } \\\\ { x = \frac { 2 \pi } { 3 } \pm 2 n \pi } \end{array}$

We know that the range of the trigonometric function is $0 \leq x < 2 \pi$ i.e., trigonometric principal solutions. This is because every solution lies between 0° and 360°

We know that the trigonometric functions tends to be periodic hence it has infinite solutions.  

For example, for $2 \cos \theta + 1$, the values of θ will be $\theta = \frac { 2 \pi } { 3 } , 2 n \pm \frac { 2 \pi } { 3 } , 4 n \pm \frac { 2 \pi } { 3 } , \ldots$  

This value will be wholly represented as $\theta = 2 n \pi \pm \frac { 2 \pi } { 3 }$ where n termed to be the integer.