Question

How do you solve $\cos \left( 2x \right)+\sin x=0$ and find all solutions in the interval $[0,2\pi )$ ?

Answer 1

The above problem is a very simple example of trigonometric equations and general values. We need to keep in mind about all the formulae and equations of trigonometry in order to smoothly solve the above given problem. In this problem, the formulae that we need to use is,
$\cos \left( 2x \right)=1-2{{\sin }^{2}}x$ . Now putting this in place of the original problem, we proceed.

Complete step by step answer:
Now, starting off with the solution to the above problem, we rewrite the problem as,
$1-2{{\sin }^{2}}x+\sin x=0$
Now, rearranging the given equation so that we form a quadratic equation in $\sin x$ , we further modify our intermediate equation as,
$2{{\sin }^{2}}x-\sin x-1=0$
Now, we can use the concept of mid-term factorization and rewrite the above equation as,
$2{{\sin }^{2}}x-2\sin x+\sin x-1=0$
Now, taking $2\sin x$ common from the first and the second term, and $1$ common from the third and the fourth terms, we get,

& 2\sin x\left( \sin x-1 \right)+1\left( \sin x-1 \right)=0 \\\ & \Rightarrow \left( \sin x-1 \right)\left( 2\sin x+1 \right)=0 \\\ \end{aligned}$$ From the above equation, we can write it in two ways, Either, $$\left( \sin x-1 \right)=0$$. Or, $$\left( 2\sin x+1 \right)=0$$ . If, $$\left( \sin x-1 \right)=0$$ , From this we can clearly write $$\sin x=1$$ . Calculating the value of $$x$$ from here, we do $$x={{\sin }^{-1}}1$$ . Now we can easily find the value of $$x$$ as, $$x=\dfrac{\pi }{2}$$ If, $$\left( 2\sin x+1 \right)=0$$ , From this we can clearly write $$\sin x=-\dfrac{1}{2}$$ . Calculating the value of $$x$$ from here, we do $$x={{\sin }^{-1}}-\dfrac{1}{2}$$ . Now we can easily find the value of $$x$$ as, $$x=\dfrac{7\pi }{6},\dfrac{11\pi }{6}$$ . **Note:** To solve these types of questions smoothly and easily, we need to keep in mind all the trigonometric general values. We must be careful to find all the values of the angles within the range which is given in the question. We can cross-check the answers by putting the values in the original equation.