# How do you solve 1+cosx-2sin^2x=0 and find all solutions in the interval 0<=x<360?

Jun 11, 2018

See below

#### Explanation:

Given $1 + \cos x - 2 {\sin}^{2} x = 0$

we can do some changes based on trigonometric identities like

$1 + \cos x - 2 \left(1 - {\cos}^{2} x\right) = 0$

$1 + \cos x - 2 + 2 {\cos}^{2} x = 0$ re-ordered we have

$2 {\cos}^{2} x + \cos x - 1 = 0$ lets be $z = \cos x$, then

$2 {z}^{2} + z - 1 = 0$ by quadratic formula

$z = \frac{- 1 \pm \sqrt{1 + 8}}{4} = \frac{- 1 \pm 3}{4}$ we obtain

${z}_{1} = - 1$ and ${z}_{2} = \frac{1}{2}$

If ${z}_{1} = \cos {x}_{1} = - 1$ then x_1=pi=180º
if ${z}_{2} = \cos {x}_{2} = \frac{1}{2}$ then x_2=pi/3=60º or x_2=300º=5pi/3

Jun 11, 2018

$x = {60}^{\circ} , {180}^{\circ} , {300}^{\circ}$

#### Explanation:

Here,

$1 + \cos x - 2 {\sin}^{2} x = 0$

$\implies 1 + \cos x - 2 \left(1 - {\cos}^{2} x\right) = 0$

$\implies 1 + \cos x - 2 + 2 {\cos}^{2} x = 0$

$\implies 2 {\cos}^{2} x + \cos x - 1 = 0$

$\implies 2 {\cos}^{2} x + 2 \cos x - \cos x - 1 = 0$

$\implies 2 \cos x \left(\cos x + 1\right) - 1 \left(\cos x + 1\right) = 0$

$\implies \left(\cos x + 1\right) \left(2 \cos x - 1\right) = 0$

$\implies \cos x + 1 = 0 \mathmr{and} 2 \cos x - 1 = 0$

$\implies \cos x = - 1 \mathmr{and} \cos x = \frac{1}{2} , w h e r e , {0}^{\circ} \le x < {360}^{\circ}$

(i)cosx=-1 < 0=>color(red)(x=180^circ

$\left(i i\right) \cos x = \frac{1}{2} > 0 \implies {I}^{s t} Q u a \mathrm{dr} a n t \mathmr{and} I {V}^{t h} Q u a \mathrm{dr} a n t$

:.I^(st)Quadrant to cosx=1/2=>color(red)(x=60^circ

IV^(th)Quadrant tocosx=1/2=>color(red)(x=360^circ-60^circ=300^circ

Hence,

$x = {60}^{\circ} , {180}^{\circ} , {300}^{\circ}$

Jun 11, 2018

$x \in \left\{60 , 180 , 300\right\}$

#### Explanation:

$\text{using the "color(blue)"trigonometric identity}$

•color(white)(x)sin^2x+cos^2x=1

$\Rightarrow {\sin}^{2} x = 1 - {\cos}^{2} x$

$1 + \cos x - 2 \left(1 - {\cos}^{2} x\right) = 0$

$2 {\cos}^{2} x + \cos x - 1 = 0 \leftarrow \textcolor{b l u e}{\text{in standard form}}$

$\left(2 \cos x - 1\right) \left(\cos x + 1\right) = 0$

$\text{equate each factor to zero and solve for x}$

$\cos x + 1 = 0 \Rightarrow \cos x = - 1 \Rightarrow x = {180}^{\circ}$

$2 \cos x - 1 = 0 \Rightarrow \cos x = \frac{1}{2}$

$\text{since "cosx>0" x in first/fourth quadrant}$

$x = {\cos}^{-} 1 \left(\frac{1}{2}\right) = {60}^{\circ} \leftarrow \textcolor{red}{\text{first quadrant}}$

$\text{or "x=(360-60)^@=300^@larrcolor(red)"fourth quadrant}$

$x \in \left\{60 , 180 , 300\right\} \to 0 \le x < 360$