4sinx×cosx+2sinx+2cosx+1=0. SOLVE.
Answers
Answer:
The value of x = n \pi \pm ( - 1 ) ^ { n + 1 } \frac { \pi } { 6 } \text { and } x = \frac { 2 \pi } { 3 } \pm 2 n \pix=nπ±(−1)n+16π and x=32π±2nπ where n terms to be the integer.
Solution:
Given,
\begin{gathered}\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}\end{gathered}4sinxcosx+2sinx+2cosx+1=02sinx(2cosx+1)+1(2cosx+1)=0(2sinx+1)(2cosx+1)=0
Either the value of x will be,
\begin{gathered}\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}\end{gathered}(2sinx+1)=0sinx=−21x=nπ±(−1)n+16π
We know that, if \sin \theta = \frac { 1 } { 2 }sinθ=21 , this is satisfied by \theta = \frac { \pi } { 6 } , \frac { 5 \pi } { 6 }θ=6π,65π etc. i.e., multiple of \frac { \pi } { 6 } \text { is } n \pi \pm \frac { \pi } { 6 }6π is nπ±6π
For infinite solutions, this value will be wholly represented as \theta = n \pi \pm ( - 1 ) ^ { n + 1 } \frac { \pi } { 6 }θ=nπ±(−1)n+16π , where n termed to be the integer.
Or the value of x will be,
\begin{gathered}\begin{array} { c } { ( 2 \cos x + 1 ) = 0 } \\\\ { \cos x = - \frac { 1 } { 2 } } \\\\ { x = \frac { 2 \pi } { 3 } \pm 2 n \pi } \end{array}\end{gathered}(2cosx+1)=0cosx=−21x=32π±2nπ
We know that the range of the trigonometric function is 0 \leq x < 2 \pi0≤x<2π 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 + 12cosθ+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 } , \ldotsθ=32π,2n±32π,4n±32π,…
This value will be wholly represented as \theta = 2 n \pi \pm \frac { 2 \pi } { 3 }θ=2nπ±32π where n termed to be the integer.