How exactly would one go about solving the following math question. I know the answer is 3 but I don't get how to arrive at this answer. A step by step explanation is greatly appreciated. Please keep in mind that I'm only a high school sophomore before you throw in any fancy trig. And i dont believe that it requires math outside of what we learnt which is pretty basic high school trig.
State the number of solutions of the equation cos(x) - sin(2x) = 1, for 0 ≤ x ≤ 2π.
cos(x) - sin(2x) = 1
=> cos(x)(1-2sin(x))=1
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$\begingroup$First try to find the solutions over $\mathbf R$, then select those which in $[0,2\pi]$.
Set $t=\tan \dfrac x 2 (x\not\equiv\pi\mod 2\pi)$. The restriction on the values of $x$ is unimportant, as one checks these values are no solutions of the equation. We know that $$\cos x=\frac{1-t^2}{1+t^2},\quad\sin x=\frac{2t}{1+t^2}.$$ Replacing in $\,\cos x(1-2\sin x)=1$ leads us, after some simplifications, to: $$2t^2(t-1)^2=0\iff \begin{cases}t=0\\t=1\end{cases}$$
Now
- $t=\tan\dfrac x2=0\iff \dfrac x2\equiv 0\mod\pi\iff x\equiv 0\mod 2\pi$,
- $t=\tan\dfrac x2=1\iff \dfrac x2\equiv \dfrac\pi 4\mod\pi\iff x\equiv \dfrac\pi 2\mod 2\pi$.
In the prescribed interval, we find only three solutions: $$x\in\Bigl\{0,\,2\pi,\,\frac\pi 2\Bigr\}$$
$\endgroup$ 7 $\begingroup$$$cos(x) - sin(2x) = 1$$ $$cos(x) - 2sin(x)cos(x) = 1$$ $$cos(x)(1 - 2sin(x)) = 1$$
If both factors in an expression multiplied are each one, then they result in one; fortunately, one solution is already found here => $$x = 0$$
Convert cos(x) into terms of sin(x)[Hint: $cos^2(x)=1- sin^2(x)$], and then find the rest of the solutions for yourself in that interval by evaluating the equation. (Evaluate it like you'd do with an algebraic expression, but now within that specific interval...)
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