Question

How do you solve the equation $2{{\left( \sin t \right)}^{2}}-\sin t-1=0$ in the interval [0,2] ? Give an answer in terms of pi ?

Answer 1

Hint: The equation given in the question is a quadratic equation where the variable is sin t . so we can find the value of sin t by using quadratic formula, if the quadratic equation is $a{{x}^{2}}+bx+c$ then the roots of the equation are $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. We can find the value of sin t by using the factorization method also.

Complete step by step answer:
The given equation is $2{{\left( \sin t \right)}^{2}}-\sin t-1=0$
If we compare the equation $2{{\left( \sin t \right)}^{2}}-\sin t-1=0$ to the quadratic equation $a{{x}^{2}}+bx+c$ where x is equal to sin t then the value of a is 2 , the value of b is -1 and the value of c is -1.
The formula for the roots of the quadratic equation is $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ , we can find the value of sin t by putting the value of a, b, and c in the above formula.
So the values of sin t are $\dfrac{-\left( -1 \right)\pm \sqrt{{{\left( -1 \right)}^{2}}-4\times 2\times \left( -1 \right)}}{2\times 2}$
Solving the equation we get
$\Rightarrow \dfrac{-\left( -1 \right)\pm \sqrt{{{\left( -1 \right)}^{2}}-4\times 2\times \left( -1 \right)}}{2\times 2}=\dfrac{1\pm \sqrt{9}}{4}$
$\Rightarrow \dfrac{-\left( -1 \right)\pm \sqrt{{{\left( -1 \right)}^{2}}-4\times 2\times \left( -1 \right)}}{2\times 2}=\dfrac{1\pm 3}{4}$
So the values of sin t are 1, $-\dfrac{1}{2}$
If the value of sin t is 1 then the only solution of t is $\dfrac{\pi }{2}$ from 0 to 2
If the value of sin t is $-\dfrac{1}{2}$ then no value of t from 0 to 2 is satisfied.

So the solution of $2{{\left( \sin t \right)}^{2}}-\sin t-1=0$ is $\dfrac{\pi }{2}$

Note: If the variable of any polynomial equation is a function of sin, cos, or exponential then we should see that the roots of the polynomial equation should come under the range of function which is variable to the polynomial function. For example if a quadratic equation has the variable as sin y then the roots of the quadratic equation should be from -1 to 1. Otherwise we will not consider that root.