
How do you solve the equation $3\sin x - 5 = 5\sin x - 4$ for $ - 360^\circ \leqslant x < 180^\circ ?$
Answer
456.9k+ views
Hint: First of all take the given expression and simplify and frame the equation in the form sine first and then in the terms of “x”. Also use All STC properties and apply for the resultant required solution.
Complete step-by-step solution:
Take the given expression: $3\sin x - 5 = 5\sin x - 4$
Move all the terms on one side of the equation. When you move any term from one side to another, then the sign of the term also changes. Positive terms become negative and vice-versa.
$3\sin x - 5 - 5\sin x + 4 = 0$
Arranged like terms together.
$\underline {3\sin x - 5\sin x} - \underline {5 + 4} = 0$
Simplify the paired like terms in the above equation. When you simplify between one negative term and one positive term you have to do subtraction and give signs of the bigger number.
$ \Rightarrow - 2\sin x - 1 = 0$
Move constant on the opposite side.
$ \Rightarrow - 2\sin x = 1$
Term multiplicative on one side of move to the opposite side then it goes to the denominator.
$ \Rightarrow \sin x = - \dfrac{1}{2}$
Referring to the All STC rule Sine is positive in the first and second quadrant and negative in third and fourth quadrant.
Therefore, here $ - 360^\circ \leqslant x < 180^\circ $so value for $x = - 30^\circ \;{\text{or x = - 150}}^\circ $
Additional Information: Remember the properties of sines and cosines and apply accordingly. The odd and even trigonometric functions states that -
$
\sin ( - \theta ) = - \sin \theta \\
\cos ( - \theta ) = \cos \theta \\
$
The most important property of sines and cosines is that their values lie between minus one and plus one. Every point on the circle is unit circle from the origin. So, the coordinates of any point are within one of zero as well.
Note: Remember the All STC rule, it is also known as ASTC rule in geometry. It states that all the trigonometric ratios in the first quadrant ($0^\circ \;{\text{to 90}}^\circ $ ) are positive, sine and cosec are positive in the second quadrant ($90^\circ {\text{ to 180}}^\circ $ ), tan and cot are positive in the third quadrant ($180^\circ \;{\text{to 270}}^\circ $ ) and sin and cosec are positive in the fourth quadrant ($270^\circ {\text{ to 360}}^\circ $ ).
Complete step-by-step solution:
Take the given expression: $3\sin x - 5 = 5\sin x - 4$
Move all the terms on one side of the equation. When you move any term from one side to another, then the sign of the term also changes. Positive terms become negative and vice-versa.
$3\sin x - 5 - 5\sin x + 4 = 0$
Arranged like terms together.
$\underline {3\sin x - 5\sin x} - \underline {5 + 4} = 0$
Simplify the paired like terms in the above equation. When you simplify between one negative term and one positive term you have to do subtraction and give signs of the bigger number.
$ \Rightarrow - 2\sin x - 1 = 0$
Move constant on the opposite side.
$ \Rightarrow - 2\sin x = 1$
Term multiplicative on one side of move to the opposite side then it goes to the denominator.
$ \Rightarrow \sin x = - \dfrac{1}{2}$
Referring to the All STC rule Sine is positive in the first and second quadrant and negative in third and fourth quadrant.
Therefore, here $ - 360^\circ \leqslant x < 180^\circ $so value for $x = - 30^\circ \;{\text{or x = - 150}}^\circ $
Additional Information: Remember the properties of sines and cosines and apply accordingly. The odd and even trigonometric functions states that -
$
\sin ( - \theta ) = - \sin \theta \\
\cos ( - \theta ) = \cos \theta \\
$
The most important property of sines and cosines is that their values lie between minus one and plus one. Every point on the circle is unit circle from the origin. So, the coordinates of any point are within one of zero as well.
Note: Remember the All STC rule, it is also known as ASTC rule in geometry. It states that all the trigonometric ratios in the first quadrant ($0^\circ \;{\text{to 90}}^\circ $ ) are positive, sine and cosec are positive in the second quadrant ($90^\circ {\text{ to 180}}^\circ $ ), tan and cot are positive in the third quadrant ($180^\circ \;{\text{to 270}}^\circ $ ) and sin and cosec are positive in the fourth quadrant ($270^\circ {\text{ to 360}}^\circ $ ).
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