Trigonometrical Equation flash cards

1. 1.State the general solution of $\sin\theta=\sin\alpha$.

   $\theta=n\pi+(-1)^n\alpha,\ n\in\mathbb{Z}$

   Hint: Alternating sign pattern with $n\pi$.

2. 2.State the general solution of $\cos\theta=\cos\alpha$.

   $\theta=2n\pi\pm\alpha,\ n\in\mathbb{Z}$

   Hint: Even multiples of $\pi$ plus or minus $\alpha$.

3. 3.State the general solution of $\tan\theta=\tan\alpha$.

   $\theta=n\pi+\alpha,\ n\in\mathbb{Z}$

   Hint: Period of $\tan$ is $\pi$, not $2\pi$.

4. 4.For what values of $k$ does $a\cos\theta+b\sin\theta=k$ have a solution?

   A solution exists iff $-\sqrt{a^2+b^2}\le k\le\sqrt{a^2+b^2}$

   Hint: Write $a\cos\theta+b\sin\theta$ as $R\sin(\theta+\phi)$ with $R=\sqrt{a^2+b^2}$.

5. 5.If $\sin\theta=0$, what is the general solution for $\theta$? Also give it for $\cos\theta=0$.

   $\sin\theta=0\Rightarrow\theta=n\pi$; $\cos\theta=0\Rightarrow\theta=(2n+1)\dfrac{\pi}{2},\ n\in\mathbb{Z}$

   Hint: Zeros of sine occur at multiples of $\pi$; zeros of cosine at odd multiples of $\pi/2$.

6. 6.Find the general solution of $\sin\theta+\sin2\theta+\sin3\theta=0$.

   Group $\sin\theta+\sin3\theta=2\sin2\theta\cos\theta$, so the equation becomes $\sin2\theta(2\cos\theta+1)=0$. From $\sin2\theta=0$ we get $\theta=\dfrac{n\pi}{2}$; from $\cos\theta=-\dfrac12$ we get $\theta=2n\pi\pm\dfrac{2\pi}{3}$. Combined: $\theta=\dfrac{n\pi}{2}$ or $\theta=2n\pi\pm\dfrac{2\pi}{3}$, $n\in\mathbb{Z}$.

   Hint: Pair the outer terms with sum-to-product first.

7. 7.Find the general solution of $\cos\theta+\cos2\theta+\cos3\theta=0$.

   Group $\cos\theta+\cos3\theta=2\cos2\theta\cos\theta$, giving $\cos2\theta(2\cos\theta+1)=0$. From $\cos2\theta=0$, $\theta=\dfrac{(2n+1)\pi}{4}$; from $\cos\theta=-\dfrac12$, $\theta=2n\pi\pm\dfrac{2\pi}{3}$. General solution: $\theta=\dfrac{(2n+1)\pi}{4}$ or $\theta=2n\pi\pm\dfrac{2\pi}{3}$.

   Hint: Same pairing trick as the sine analogue — factor out $\cos2\theta$.

8. 8.Solve $\tan3x=\tan x$ for the general solution, being careful to reject any values excluded by the domain of $\tan$.

   Naively $3x=n\pi+x\Rightarrow x=\dfrac{n\pi}{2}$. But $\tan x$ is undefined when $x=\dfrac{\pi}{2}+k\pi$, which occurs exactly for odd $n$ in this family. Checking, $x=\dfrac{n\pi}{2}$ with $n$ odd must be discarded, while $3x$ never becomes undefined at the surviving points. Final general solution: $x=n\pi$, $n\in\mathbb{Z}$ (only even $n$ survive).

   Hint: Solve first, then test which family members keep tan x defined.

9. 9.Find the general solution of $2\sin^2x+\sin x-1=0$.

   Let $t=\sin x$: $2t^2+t-1=(2t-1)(t+1)=0\Rightarrow t=\dfrac12$ or $t=-1$; both lie in $[-1,1]$ so neither is spurious. $\sin x=\dfrac12\Rightarrow x=n\pi+(-1)^n\dfrac{\pi}{6}$; $\sin x=-1\Rightarrow x=2n\pi-\dfrac{\pi}{2}$. Both families are valid.

   Hint: Factor the quadratic in sin x, then check both roots lie in [-1,1].

10. 10.Solve $\sin x+\cos x=1$ for $x\in[0,2\pi)$ by squaring both sides, and identify which of the resulting candidates are extraneous.

    Squaring gives $1+\sin2x=1\Rightarrow\sin2x=0\Rightarrow x=0,\dfrac{\pi}{2},\pi,\dfrac{3\pi}{2}$. Testing each in the original equation: $x=0$ and $x=\dfrac{\pi}{2}$ satisfy it, but $x=\pi$ and $x=\dfrac{3\pi}{2}$ give $\sin x+\cos x=-1$, so they are extraneous roots introduced by squaring. Valid solutions: $x=0,\dfrac{\pi}{2}$ (general form $x=2n\pi$ or $x=2n\pi+\dfrac{\pi}{2}$).

    Hint: Squaring can flip a sign — verify every candidate in the original equation.

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