Answer
452.7k+ views
Hint: Since, we have different terms (in terms that on LHS, there are trigonometric terms containing $\theta $ and on RHS, there is a $2\theta $ term) on LHS and RHS of the trigonometric equation, we solve this trigonometric equation by squaring LHS and RHS terms. This way we will be able to simplify the equation easily. Further, it is also useful to know the formula of sin$2\theta $ in terms of sin$\theta $ and cos$\theta $. This is given by - sin$2\theta $=2 sin$\theta $cos$\theta $
Complete step-by-step solution:
Thus, we have,
$\sin \theta +\cos \theta =\sin 2\theta $
Now, squaring the LHS and RHS, we have,
${{(\sin \theta +\cos \theta )}^{2}}={{(\sin 2\theta )}^{2}}$
\[{{(\sin \theta )}^{2}}+{{(\cos \theta )}^{2}}+2\sin \theta \cos \theta ={{(\sin 2\theta )}^{2}}\]
1+\[\sin 2\theta ={{(\sin 2\theta )}^{2}}\] -- (1)
Since, \[{{(\sin \theta )}^{2}}+{{(\cos \theta )}^{2}}=1\] and \[2\sin \theta \cos \theta =\sin 2\theta \]
Now, solving (1) further,
1+\[\sin 2\theta ={{(\sin 2\theta )}^{2}}\]
Let \[\sin 2\theta \]=t, thus, we have,
1+$t={{t}^{2}}$
${{t}^{2}}-t-1=0$-- (2)
To solve, $a{{x}^{2}}+bx+c=0$, the solution is-
x=$\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Now solving the equation (2), we have,
t= $\dfrac{1\pm \sqrt{{{1}^{2}}-4(-1)(1)}}{2(1)}$
t=$\dfrac{1\pm \sqrt{5}}{2}$
Now, we can eliminate, t=$\dfrac{1+\sqrt{5}}{2}$=1.618
Since, t=$\sin 2\theta $ would be greater than 1. This would not be possible since, -1$\le $$\sin 2\theta $$\le $1.
Thus, we only have one solution,
t=$\dfrac{1-\sqrt{5}}{2}$
Since, t=$\sin 2\theta $, we have,
$\sin 2\theta $=$\dfrac{1-\sqrt{5}}{2}$
Now, to solve this equation, we will make use of graph,
Now, to get the number of solutions, we have to find the number of intersection of y=$\dfrac{1-\sqrt{5}}{2}$and y=$\sin 2\theta $ between [$\pi ,-\pi $]. (Also, for reference, $\pi $ is approximately 3.14)
Clearly, we see that there are four intersections within the range [$\pi ,-\pi $].
Hence, there are four solutions to the trigonometric equation $\sin \theta +\cos \theta =\sin 2\theta $.
Note: While solving trigonometric equations, we should try to solve the question by bringing LHS and RHS in same degree of angle (that is, in this case, by squaring the LHS and RHS terms, we were eventually able to bring both LHS and RHS in terms of $2\theta $. Finally, we should ensure that the solution is always within the limits [-1,1] since, -1$\le $$\sin x$$\le $1.
Complete step-by-step solution:
Thus, we have,
$\sin \theta +\cos \theta =\sin 2\theta $
Now, squaring the LHS and RHS, we have,
${{(\sin \theta +\cos \theta )}^{2}}={{(\sin 2\theta )}^{2}}$
\[{{(\sin \theta )}^{2}}+{{(\cos \theta )}^{2}}+2\sin \theta \cos \theta ={{(\sin 2\theta )}^{2}}\]
1+\[\sin 2\theta ={{(\sin 2\theta )}^{2}}\] -- (1)
Since, \[{{(\sin \theta )}^{2}}+{{(\cos \theta )}^{2}}=1\] and \[2\sin \theta \cos \theta =\sin 2\theta \]
Now, solving (1) further,
1+\[\sin 2\theta ={{(\sin 2\theta )}^{2}}\]
Let \[\sin 2\theta \]=t, thus, we have,
1+$t={{t}^{2}}$
${{t}^{2}}-t-1=0$-- (2)
To solve, $a{{x}^{2}}+bx+c=0$, the solution is-
x=$\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Now solving the equation (2), we have,
t= $\dfrac{1\pm \sqrt{{{1}^{2}}-4(-1)(1)}}{2(1)}$
t=$\dfrac{1\pm \sqrt{5}}{2}$
Now, we can eliminate, t=$\dfrac{1+\sqrt{5}}{2}$=1.618
Since, t=$\sin 2\theta $ would be greater than 1. This would not be possible since, -1$\le $$\sin 2\theta $$\le $1.
Thus, we only have one solution,
t=$\dfrac{1-\sqrt{5}}{2}$
Since, t=$\sin 2\theta $, we have,
$\sin 2\theta $=$\dfrac{1-\sqrt{5}}{2}$
Now, to solve this equation, we will make use of graph,
Now, to get the number of solutions, we have to find the number of intersection of y=$\dfrac{1-\sqrt{5}}{2}$and y=$\sin 2\theta $ between [$\pi ,-\pi $]. (Also, for reference, $\pi $ is approximately 3.14)
Clearly, we see that there are four intersections within the range [$\pi ,-\pi $].
Hence, there are four solutions to the trigonometric equation $\sin \theta +\cos \theta =\sin 2\theta $.
Note: While solving trigonometric equations, we should try to solve the question by bringing LHS and RHS in same degree of angle (that is, in this case, by squaring the LHS and RHS terms, we were eventually able to bring both LHS and RHS in terms of $2\theta $. Finally, we should ensure that the solution is always within the limits [-1,1] since, -1$\le $$\sin x$$\le $1.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)