Answer
Verified
429k+ views
Hint: Here we will use the sine angle identity will replace the formula and find the factors for the given equation. Then we will use the All STC rule to find the required resultant angles.
Complete step-by-step solution:
Take the given equation: $ \sin 4x + \sin 2x = 0 $
The above equation can be re-written as: $ \sin 2(2x) + \sin 2x = 0 $
Using the identity: $ \sin 2\theta = 2\sin \theta \cos \theta $ in the above equation.
$ \Rightarrow 2\sin 2x\cos 2x + \sin 2x = 0 $
Take common multiples from the above equation.
$ \Rightarrow \sin 2x(2\cos 2x + 1) = 0 $
From the above equation there are two cases:
Case (I)
$ \Rightarrow \sin 2x = 0 $
Take the inverse of sine on both the sides of the above equation.
$ \Rightarrow {\sin ^{ - 1}}(\sin 2x) = {\sin ^{ - 1}}(0) $
Referring to the trigonometric table for sine value,
$ \Rightarrow {\sin ^{ - 1}}(\sin 2x) = {\sin ^{ - 1}}(\sin 0^\circ ) $
Sine inverse and sine cancel each other.
$ \Rightarrow 2x = 0 $
Make “x” has the subject
$ \Rightarrow x = \dfrac{0}{2} $
Zero upon anything is always zero.
$ \Rightarrow x = 0 $
We know that in a circle for every $ 360^\circ $ angle all values are repeated.
$ \therefore x = 0,2\pi ,4\pi ,..... $ ….(A)
Case (II)
$ 2\cos 2x + 1 = 0 $
Make the subject angle and move other terms on the opposite side. Remember when you move any term from one to another, the sign of the term also changes. Positive terms become negative.
$ \Rightarrow 2\cos 2x = - 1 $
Term multiplicative on one side if moved to the opposite side, then it goes to the denominator.
$ \Rightarrow \cos 2x = - \dfrac{1}{2} $
Cosine is negative in the second and third quadrant.
$ \Rightarrow 2x = \dfrac{{2\pi }}{3},\dfrac{{4\pi }}{3},\dfrac{{8\pi }}{3},..... $
Simplify the above equation,
$ \Rightarrow x = \dfrac{\pi }{3},\dfrac{{2\pi }}{3},\dfrac{{4\pi }}{3},..... $ ….(B)
Equations (A) and (B) are the required solution.
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 equation: $ \sin 4x + \sin 2x = 0 $
The above equation can be re-written as: $ \sin 2(2x) + \sin 2x = 0 $
Using the identity: $ \sin 2\theta = 2\sin \theta \cos \theta $ in the above equation.
$ \Rightarrow 2\sin 2x\cos 2x + \sin 2x = 0 $
Take common multiples from the above equation.
$ \Rightarrow \sin 2x(2\cos 2x + 1) = 0 $
From the above equation there are two cases:
Case (I)
$ \Rightarrow \sin 2x = 0 $
Take the inverse of sine on both the sides of the above equation.
$ \Rightarrow {\sin ^{ - 1}}(\sin 2x) = {\sin ^{ - 1}}(0) $
Referring to the trigonometric table for sine value,
$ \Rightarrow {\sin ^{ - 1}}(\sin 2x) = {\sin ^{ - 1}}(\sin 0^\circ ) $
Sine inverse and sine cancel each other.
$ \Rightarrow 2x = 0 $
Make “x” has the subject
$ \Rightarrow x = \dfrac{0}{2} $
Zero upon anything is always zero.
$ \Rightarrow x = 0 $
We know that in a circle for every $ 360^\circ $ angle all values are repeated.
$ \therefore x = 0,2\pi ,4\pi ,..... $ ….(A)
Case (II)
$ 2\cos 2x + 1 = 0 $
Make the subject angle and move other terms on the opposite side. Remember when you move any term from one to another, the sign of the term also changes. Positive terms become negative.
$ \Rightarrow 2\cos 2x = - 1 $
Term multiplicative on one side if moved to the opposite side, then it goes to the denominator.
$ \Rightarrow \cos 2x = - \dfrac{1}{2} $
Cosine is negative in the second and third quadrant.
$ \Rightarrow 2x = \dfrac{{2\pi }}{3},\dfrac{{4\pi }}{3},\dfrac{{8\pi }}{3},..... $
Simplify the above equation,
$ \Rightarrow x = \dfrac{\pi }{3},\dfrac{{2\pi }}{3},\dfrac{{4\pi }}{3},..... $ ….(B)
Equations (A) and (B) are the required solution.
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 $ ).
Recently Updated Pages
10 Examples of Evaporation in Daily Life with Explanations
10 Examples of Diffusion in Everyday Life
1 g of dry green algae absorb 47 times 10 3 moles of class 11 chemistry CBSE
If the coordinates of the points A B and C be 443 23 class 10 maths JEE_Main
If the mean of the set of numbers x1x2xn is bar x then class 10 maths JEE_Main
What is the meaning of celestial class 10 social science CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
In the tincture of iodine which is solute and solv class 11 chemistry CBSE
Why is there a time difference of about 5 hours between class 10 social science CBSE