Question

The number of distinct solutions of the equation $\dfrac{5}{4}{{\cos }^{2}}2x+{{\cos }^{4}}x+{{\sin }^{4}}x+{{\cos }^{6}}x+{{\sin }^{6}}x=2$ in the interval $\left[ 0,2\pi \right]$ is____________.

Answer 1

Hint: To solve the above question, we need the trigonometric formulas like ${{\cos }^{2}}x+{{\sin }^{2}}x=1$and then most important formulas ${{\cos }^{4}}x+{{\sin }^{4}}x={{\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)}^{2}}-2{{\sin }^{2}}x.{{\cos }^{2}}x$ we will apply this in the given question and ${{\cos }^{6}}x+{{\sin }^{6}}x=\left( {{\cos }^{2}}x+{{\sin }^{2}}x \right)\left( {{\cos }^{4}}x+{{\sin }^{4}}x-2{{\sin }^{2}}x{{\cos }^{2}}x \right)$with the help of these formulas we will solve the given question.
we have to know the formula of ${{\cos }^{2}}x+{{\sin }^{2}}x=1$ .So, first we have to use the formula of ${{\cos }^{2}}x+{{\sin }^{2}}x=1$ .By the time of determine the solutions we have to keep in mind that we have to determine the solution which is under $\left[ 0,2\pi \right]$ .

Complete step-by-step answer:
Now it is give that the equation as,
$\dfrac{5}{4}{{\cos }^{2}}2x+{{\cos }^{4}}x+{{\sin }^{4}}x+{{\cos }^{6}}x+{{\sin }^{6}}x=2$
First of all, we simplify the ${{\cos }^{4}}x+{{\sin }^{4}}x$
We can see that it is of formula of ${{a}^{2}}+{{b}^{2}}$ that is ${{\left( a+b \right)}^{2}}-2ab$ .
So we just simplify it and we will get,
$\left( {{\cos }^{4}}x+{{\sin }^{4}}x \right)={{\left( {{\cos }^{2}}x \right)}^{2}}+{{\left( {{\sin }^{2}}x \right)}^{2}}$
Now we make it simplify and we will get,
${{\left( {{\cos }^{2}}x \right)}^{2}}+{{\left( {{\sin }^{2}}x \right)}^{2}}={{\left( {{\cos }^{2}}x+{{\sin }^{2}}x \right)}^{2}}-2{{\sin }^{2}}x{{\cos }^{2}}x$
And we know that ${{\cos }^{2}}x+{{\sin }^{2}}x=1$
Now we just use the formula and simplify it and we will get,
$\Rightarrow 1-2{{\sin }^{2}}x{{\cos }^{2}}x$
Now we can simplify the formula ${{\cos }^{6}}x+{{\sin }^{6}}x$ .
We can see that it is of formula of ${{a}^{3}}+{{b}^{3}}$ that is $\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)$ .
So, now we make it simplify and we will get,
${{\cos }^{6}}x+{{\sin }^{6}}x={{\left( {{\cos }^{2}}x \right)}^{3}}+{{\left( {{\sin }^{2}}x \right)}^{3}}$
Now we just use the formula and simplify it and we will get,
${{\left( {{\cos }^{2}}x \right)}^{3}}+{{\left( {{\sin }^{2}}x \right)}^{3}}=\left( {{\cos }^{2}}x+{{\sin }^{2}}x \right)\left( {{\cos }^{4}}x+{{\sin }^{4}}x-2{{\sin }^{2}}x{{\cos }^{2}}x \right)$
By the time of simplifying of $\left( {{\cos }^{4}}x+{{\sin }^{4}}x \right)$ , we can get
$\left( {{\cos }^{4}}x+{{\sin }^{4}}x \right)=1-2{{\sin }^{2}}x{{\cos }^{2}}x$
And we know that ${{\cos }^{2}}x+{{\sin }^{2}}x=1$
So, we can get simplifying the formula is
${{\cos }^{6}}x+{{\sin }^{6}}x=1\times \left( 1-2{{\sin }^{2}}x{{\cos }^{2}}x-{{\sin }^{2}}x{{\cos }^{2}}x \right)$
After simplifying it we get,
${{\cos }^{6}}x+{{\sin }^{6}}x=\left( 1-3{{\sin }^{2}}x{{\cos }^{2}}x \right)$
Now we come to our problem, we will get,
$\dfrac{5}{4}{{\cos }^{2}}2x+{{\cos }^{4}}x+{{\sin }^{4}}x+{{\cos }^{6}}x+{{\sin }^{6}}x=2$ \[\Rightarrow \dfrac{5}{4}{{\cos }^{2}}2x+1-2{{\sin }^{2}}x{{\cos }^{2}}x+\left( 1-3{{\sin }^{2}}x{{\cos }^{2}}x \right)\]
Now we will simplify it and we will get,
$\Rightarrow \dfrac{5}{4}{{\cos }^{2}}2x-5{{\sin }^{2}}x{{\cos }^{2}}x=0$
Now we will use the formula ${{\cos }^{2}}2x=1-{{\sin }^{2}}2x$
So, we will get,
$\dfrac{5}{4}-\dfrac{5}{4}{{\sin }^{2}}2x-\dfrac{5}{4}{{\sin }^{2}}2x=0$
Where we use the formula $\sin 2x=2\sin x\cos x$
By simplifying it we will get,
${{\sin }^{2}}2x=\dfrac{1}{2}$
Both sides doing square root,
$\Rightarrow \sin 2x=\pm \dfrac{1}{\sqrt{2}}$
Now we will get,
$\therefore x=\dfrac{\pi }{8},\dfrac{3\pi }{8},\dfrac{5\pi }{8},\dfrac{7\pi }{8},\dfrac{9\pi }{8},\dfrac{11\pi }{8},\dfrac{13\pi }{8},\dfrac{15\pi }{8}$
Hence the number of solutions in the given interval is $8$ .

Note: For such kind of questions you need to know the formulas all trigonometric formulas mostly the students are confused at the first site while exam it is very simple to solve such problem first of all you need to simplify the trigonometric term and then using the relation ${{\cos }^{2}}x+{{\sin }^{2}}x=1$ just simplify the given terms that will definitely get you the answer. And also, mostly higher the power higher is chance of getting wrong so there are couple of basic formulas like${{\cos }^{4}}x+{{\sin }^{4}}x={{\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)}^{2}}-2{{\sin }^{2}}x.{{\cos }^{2}}x$ and ${{\cos }^{6}}x+{{\sin }^{6}}x=\left( {{\cos }^{2}}x+{{\sin }^{2}}x \right)\left( {{\cos }^{4}}x+{{\sin }^{4}}x-2{{\sin }^{2}}x{{\cos }^{2}}x \right)$
These are the formulas for the higher order that will help you during the problem solving remember it if you don’t know.