Question

Determine all solutions to the equation 2sin2 x = 1 + cos x on the interval [0, 2π).

Answer 1

`x=(\pi)/(3),(5\pi)/(3),\pi`

Step-by-step explanation:

Given the below;

`2\sin ^2x=1+\cos x`

We'll follow the below steps to solve for x;

Step 1: Rewrite using the below trig identity;

`\begin{gathered} \sin ^2x+\cos ^2x=1 \\ \therefore\sin ^2x=1-\cos ^2x \end{gathered}`

So, we'll have;

`\begin{gathered} 2(1-\cos ^2x)=1+\cos x \\ \end{gathered}`

Step 2: Apply the difference of squares formula to the left-hand side of the equation;

`\begin{gathered} 2(1^2-\cos ^2x)=1+\cos x \\ 2(1+\cos x)(1-\cos x)=1+\cos x \end{gathered}`

Step 3: Subtract (1 + cos x) from both sides;

`2(1+\cos x)(1-\cos x)-(1+\cos x)=0`

Step 4: Factor out (1 + cos x);

`\begin{gathered} (1+\cos x)\lbrack2(1-\cos x)-1\rbrack=0 \\ (1+\cos x)(2-2\cos x-1)=0_{} \\ (1+\cos x)(1-2\cos x)=0 \end{gathered}`

Step 5: Solve for the values of x by equating each term to zero;

`\begin{gathered} 1+\cos x=0 \\ \cos x=-1 \\ x=\cos ^(-1)(-1) \\ x=180^(\circ) \\ \therefore x=\pi \\ Or \\ 1-2\cos x=0 \\ -2\cos x=-1 \\ (-2\cos x)/(-2)=(-1)/(-2) \\ \cos x=(1)/(2) \\ x=\cos ^(-1)((1)/(2)) \\ x=60^(\circ),300^(\circ) \\ \therefore x=(\pi)/(3),(5\pi)/(3) \end{gathered}`

`\begin{gathered} \\ \end{gathered}`