Question

How do you solve $3\cos x+3=2{{\sin }^{2}}x$ over the interval 0 to $2\pi$?

Answer 1

To solve this problem, we should know some of the properties and rules of trigonometry and algebra as follows, we should know the trigonometric identity which states ${{\sin }^{2}}x+{{\cos }^{2}}x=1$. The range of cosine functions is always $[-1,1]$. For a quadratic equation $a{{x}^{2}}+bx+c=0$, we can find the roots of the equation using the formula method as, $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. We can find the roots by substituting the value of coefficients.

Complete step by step solution:
We are asked to solve the equation $3\cos x+3=2{{\sin }^{2}}x$. We know the trigonometric identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$, using this we can simplify the equation as,
$\Rightarrow 3\cos x+3=2\left( 1-{{\cos }^{2}}x \right)$
Expanding the bracket on the right-hand side using distributive property, we get
$\Rightarrow 3\cos x+3=2-2{{\cos }^{2}}x$
Simplifying the above equation, it can be expressed as
$\Rightarrow 2{{\cos }^{2}}x+3\cos x+1=0$
Substituting $\cos x=t$ in the above equation, we get
$\Rightarrow 2{{t}^{2}}+3t+1=0$
The above equation is a quadratic in t. compared with the general quadratic equation $a{{x}^{2}}+bx+c=0$, we get $a=2,b=3\And c=1$. We can find the roots of the equation using the formula method $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. Substituting the values of the coefficients of the equation, we get

& \Rightarrow t=\dfrac{-3\pm \sqrt{{{3}^{2}}-4(2)(1)}}{2(2)} \\\ & \Rightarrow t=\dfrac{-3\pm \sqrt{1}}{4} \\\ & \Rightarrow t=\dfrac{-3\pm 1}{4} \\\ \end{aligned}$$ $$\Rightarrow t=\dfrac{-3+1}{4}$$ or $$t=\dfrac{-3-1}{4}$$ $$\therefore t=\dfrac{-1}{2}$$ or $$t=-1$$ Using the substitution, we get $$\cos x=\dfrac{-1}{2}$$ or $$\cos x=-1$$. In the range of $$\left[ 0,2\pi \right]$$, we know that two angles are satisfying $$\cos x=\dfrac{-1}{2}$$ they are $$\dfrac{2\pi }{3}\And \dfrac{4\pi }{3}$$. Similarly, for $$\cos x=-1$$, there is one angle in the range $$\left[ 0,2\pi \right]$$ it is $$\pi $$. Thus, the solutions for the given equations are $$\pi ,\dfrac{2\pi }{3}\And \dfrac{4\pi }{3}$$. **Note:** For questions having a polynomial in terms of trigonometric functions. We must check if the solution lies in the range of trigonometric ratios or not. If the solution does not lie, then exclude the solution. This problem does not arise for tangent and cotangent ratios as their range is real numbers.