Solving $a\cos x + b\sin x = c$
You can convert $a\cos x + b\sin x$ to $R\cos(x-\alpha)$. Now add a right-hand side $c$ and the equation becomes $R\cos(x-\alpha)=c$ — a standard cosine equation you already know how to solve. This lesson drills the three-step pipeline: convert, divide, solve. Master it here and you'll handle every variant the exam throws at you.
Consider the equation $\sqrt{3}\cos x + \sin x = 1$ on $[0, 2\pi]$. Before solving it — how many solutions do you think there are, and why? Write your reasoning.
Solving $a\cos x + b\sin x = c$ always follows three steps. Once you internalise the pipeline, every problem of this type becomes a two-minute routine.
Step 1 — Convert. Write $a\cos x + b\sin x = R\cos(x-\alpha)$ using $R = \sqrt{a^2+b^2}$ and $\tan\alpha = b/a$ (correct quadrant).
Step 2 — Divide. The equation becomes $\cos(x-\alpha) = \dfrac{c}{R}$. This has solutions only if $\left|\dfrac{c}{R}\right| \leq 1$, i.e. $|c| \leq R$.
Step 3 — Solve. Let $u = x - \alpha$. Solve $\cos u = c/R$ over the translated domain, then back-substitute $x = u + \alpha$.
Key facts
- $a\cos x + b\sin x = c$ transforms to $\cos(x-\alpha) = c/R$ after converting and dividing
- Solutions exist only when $|c| \leq R = \sqrt{a^2+b^2}$
- General solution: $x = \alpha \pm \cos^{-1}(c/R) + 2n\pi$, $n \in \mathbb{Z}$
Concepts
- Why the domain must be translated when substituting $u = x - \alpha$
- How the $\pm$ in the cosine solution captures both intersection points of the horizontal line $y = c/R$ with the cosine curve
- Why the condition $|c| \leq R$ corresponds geometrically to the line $y = c/R$ intersecting $y = \cos\theta$
Skills
- Solve $a\cos x + b\sin x = c$ fully in a specified domain, showing all working
- Identify when an equation has no solutions, one solution (tangent case), or two solutions per period
- Write the general solution when no domain is specified
To solve $a\cos x + b\sin x = c$ on a domain such as $[0, 2\pi]$:
Step 1 — Convert
$a\cos x + b\sin x = R\cos(x-\alpha)$ where $R = \sqrt{a^2+b^2}$, $\cos\alpha = a/R$, $\sin\alpha = b/R$.
Step 2 — Check and simplify
Check $|c| \leq R$ (no solutions if violated). Equation becomes $\cos(x-\alpha) = \dfrac{c}{R}$.
Step 3 — Let $u = x - \alpha$
Translate domain: if $x\in[0,2\pi]$ then $u\in[-\alpha, 2\pi-\alpha]$.
Solve $\cos u = c/R$: $u = \beta$ or $u = -\beta$ (where $\beta = \cos^{-1}(c/R)$), adjusted for the translated domain.
Step 4 — Back-substitute
Pipeline: convert check |c| R divide let u=x- translate domain solve back-substitute; Two solutions per period when |c/R| 1
Pause — copy the full solution pipeline into your book: convert to $R\sin(x-\alpha)$ → check $|c| \le R$ → let $u = x-\alpha$ → translate domain → solve $\sin u = c/R$ → back-substitute $x = u + \alpha$.
Quick check: For which value of $c$ does $3\cos x + 4\sin x = c$ have no solutions?
We just saw the seven-step pipeline ending with the substitution $u = x-\alpha$ to reduce to a standard $\sin u = k$ equation. That raises a question: when you substitute $u = x - \alpha$, the original domain $[0, 2\pi]$ for $x$ shifts — what becomes the new domain for $u$? This card answers it → the domain for $u$ is $[-\alpha, 2\pi-\alpha]$; cosine's even symmetry means negative $u$ values still yield valid solutions.
The most common error after substituting $u = x - \alpha$ is forgetting to translate the domain. The original constraint is on $x$, but now you are solving for $u$.
If $x \in [0, 2\pi]$ and $\alpha = \pi/6$, then:
$u = x - \alpha \in [0-\pi/6,\; 2\pi-\pi/6] = [-\pi/6,\; 11\pi/6]$
Solve $\cos u = c/R$ within this translated interval, listing all solutions for $u$, then add $\alpha = \pi/6$ to each to recover $x$.
u = x - : domain shifts from [0,2] to [-, 2 - ]; Cosine is even: (-u) = u, so negative u values are valid
Pause — copy the domain-translation rule into your book: $u = x - \alpha$ shifts the domain from $[0,2\pi]$ to $[-\alpha, 2\pi-\alpha]$; cosine is even so $\cos(-u) = \cos u$, which may yield extra valid solutions.
Did you get this? True or false: if $\alpha = \pi/3$ and $x \in [0, 2\pi]$, the translated domain for $u = x - \alpha$ is $[-\pi/3,\; 5\pi/3]$.
Worked examples · 3 in a row, reveal as you go
Solve $\sqrt{3}\cos x + \sin x = 1$ for $x \in [0, 2\pi]$.
Solve $3\cos x + 4\sin x = 6$ for $x \in [0, 2\pi]$.
Find the general solution of $\cos x - \sin x = 1$.
Fill the gap: For $3\cos x + 4\sin x = c$, solutions exist only when $|c| \leq$ .
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: the equation $2\cos x - 2\sin x = 3$ has no solutions because the amplitude $R = 2\sqrt{2} < 3$.
Activities · practice with the ideas
Solve $3\cos x + 4\sin x = 4$ for $x \in [0, 2\pi]$. Show all three steps.
Determine whether $\cos x + \sin x = \sqrt{3}$ has solutions on $[0, 2\pi]$. If so, find them; if not, explain why.
Solve $-\cos x + \sqrt{3}\sin x = 1$ for $x \in [0, 2\pi]$. (Hint: $\alpha$ is in Q2.)
Find the general solution of $\sqrt{3}\cos x - \sin x = \sqrt{3}$.
For $a\cos x + b\sin x = c$ with $R = \sqrt{a^2+b^2}$, show that the equation has exactly one solution per period when $|c| = R$, and find that solution (in terms of $\alpha$).
Odd one out: Three of these statements about $a\cos x + b\sin x = c$ are correct. Which one is WRONG?
Earlier you guessed the number of solutions to $\sqrt{3}\cos x + \sin x = 1$ on $[0, 2\pi]$.
The exact solutions are $x = \pi/2$ and $x = 11\pi/6$ — two solutions. This is typical whenever $|c/R| < 1$: the horizontal line $y = c/R$ cuts the cosine curve in two points per period. If you guessed two, your geometric intuition is already working. Reflect on the step that was hardest for you in the worked example.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. State the values of $c$ for which $\cos x + \sqrt{3}\sin x = c$ has solutions, and find $R$. (2 marks)
Q2. Solve $\cos x + \sqrt{3}\sin x = \sqrt{3}$ for $x \in [0, 2\pi]$. (3 marks)
Q3. Find all solutions of $3\cos x + 4\sin x = 5$ in $[0, 2\pi]$. Explain why there is only one solution. (3 marks)
Comprehensive answers (click to reveal)
Activity answers:
1. $3\cos x+4\sin x=4$: $R=5$, $\alpha=\arctan(4/3)$. $\cos(x-\alpha)=4/5$. Let $u=x-\alpha$: $u=\cos^{-1}(4/5)\approx0.6435$ or $u\approx-0.6435$ (equiv. $\approx5.640$ in $[0,2\pi]$). $x=\alpha+0.6435\approx 1.495$ and $x=\alpha+5.640\approx6.57$ — check domain; adjust for translated domain $[-\alpha, 2\pi-\alpha]$.
2. $\cos x+\sin x=\sqrt{3}$: $R=\sqrt{2}\approx1.414 < \sqrt{3}\approx1.732$. Since $|c|>R$, no solutions.
3. $-\cos x+\sqrt{3}\sin x=1$: $R=2$, $\alpha=2\pi/3$. $\cos(x-2\pi/3)=1/2$. Translated domain $u\in[-2\pi/3, 4\pi/3]$. $u=\pi/3$ or $u=-\pi/3$. $x=\pi/3+2\pi/3=\pi$ and $x=-\pi/3+2\pi/3=\pi/3$.
4. $\sqrt{3}\cos x-\sin x=\sqrt{3}$: $R=2$, $\alpha=-\pi/6$. $\cos(x+\pi/6)=\sqrt{3}/2$. $u=\pm\pi/6+2n\pi$. $x=0+2n\pi$ or $x=-\pi/3+2n\pi$.
5. When $|c|=R$: $\cos(x-\alpha)=\pm1$. $+1$ gives $x-\alpha=2n\pi$, i.e. $x=\alpha+2n\pi$ (one solution per period); $-1$ gives $x=\alpha+\pi+2n\pi$ (another set). So one solution per $2\pi$-period from each case.
Q1 (2 marks): $R = \sqrt{1+3} = 2$ [1]. Solutions exist when $|c| \leq 2$, i.e. $-2 \leq c \leq 2$ [1].
Q2 (3 marks): $R=2$, $\alpha=\pi/3$: $\cos(x-\pi/3) = \sqrt{3}/2$ [1]. Translated domain: $u\in[-\pi/3, 5\pi/3]$. $\cos u=\sqrt{3}/2 \Rightarrow u=\pi/6$ or $u=-\pi/6$. Both in $[-\pi/3, 5\pi/3]$ ✓ [1]. $x=\pi/6+\pi/3=\pi/2$ and $x=-\pi/6+\pi/3=\pi/6$ [1].
Q3 (3 marks): $R=5$, $|c/R|=5/5=1$ — tangent case, so exactly one solution per period [1]. $\cos(x-\alpha)=1 \Rightarrow x-\alpha=0 \Rightarrow x=\alpha=\tan^{-1}(4/3)\approx0.927$ rad [1]. Check: $3\cos(0.927)+4\sin(0.927)=3(0.6)+4(0.8)=1.8+3.2=5$ ✓ [1].
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