Here are the solutions for the given trigonometric equations:
5a) Solve cos(θ+30∘)=sinθ for 0∘≤θ<360∘.
Step 1: Expand cos(θ+30∘) using the compound angle formula cos(A+B)=cosAcosB−sinAsinB.
cos(θ+30∘)=cosθcos30∘−sinθsin30∘
Substitute the known values cos30∘=23 and sin30∘=21:
cos(θ+30∘)=23cosθ−21sinθ
Step 2: Substitute this back into the original equation and rearrange to solve for tanθ.
23cosθ−21sinθ=sinθ23cosθ=sinθ+21sinθ23cosθ=23sinθ
Divide both sides by cosθ and 23:
33=cosθsinθtanθ=33
Step 3: Find the reference angle and solutions in the given domain.
The reference angle for tanθ=33 is 30∘.
Since tanθ is positive, θ lies in Quadrant I and Quadrant III.
In Quadrant I: θ=30∘.
In Quadrant III: θ=180∘+30∘=210∘.
The solutions are θ=30∘,210∘.
5b) Solve 2cos(θ−45∘)=4sinθ for 0∘≤θ<360∘.
Step 1: Expand cos(θ−45∘) using the compound angle formula cos(A−B)=cosAcosB+sinAsinB.
cos(θ−45∘)=cosθcos45∘+sinθsin45∘
Substitute the known values cos45∘=22 and sin45∘=22:
cos(θ−45∘)=22cosθ+22sinθ
Step 2: Substitute this back into the original equation and simplify.
2(22cosθ+22sinθ)=4sinθ(22cosθ+22sinθ)=4sinθcosθ+sinθ=4sinθ
Step 3: Rearrange the equation to solve for tanθ.
cosθ=4sinθ−sinθcosθ=3sinθ
Divide both sides by cosθ:
1=3cosθsinθtanθ=31
Step 4: Find the reference angle and solutions in the given domain.
The reference angle for tanθ=31 is arctan(31)≈18.43∘.
Since tanθ is positive, θ lies in Quadrant I and Quadrant III.
In Quadrant I: θ≈18.43∘.
In Quadrant III: θ=180∘+18.43∘=198.43∘.
The solutions are θ≈18.43∘,198.43∘.
6a) Solve sin50∘cosx+cos50∘sinx=1 for 0∘≤x<360∘.
Step 1: Recognize the compound angle formula sin(A+B)=sinAcosB+cosAsinB.
The left side of the equation matches this form with A=50∘ and B=x.
So, the equation can be rewritten as:
sin(50∘+x)=1
Step 2: Find the general solution for the angle (50∘+x).
The sine function is equal to 1 when the angle is 90∘ plus any multiple of 360∘.
50∘+x=90∘+n⋅360∘
Step 3: Solve for x in the domain 0∘≤x<360∘.
x=90∘−50∘+n⋅360∘x=40∘+n⋅360∘
For n=0: x=40∘.
For other integer values of n, x will be outside the given domain.
The solution is x=40∘.
6b) Solve tanx−tan35∘=1+tanxtan35∘ for 0∘≤x<360∘.
Step 1: Rearrange the equation to match the compound angle formula for tan(A−B)=1+tanAtanBtanA−tanB.
Divide both sides by (1+tanxtan35∘):
1+tanxtan35∘tanx−tan35∘=1
Step 2: Recognize the left side as tan(x−35∘).
So, the equation becomes:
tan(x−35∘)=1
Step 3: Find the general solution for the angle (x−35∘).
The tangent function is equal to 1 when the angle is 45∘ plus any multiple of 180∘.
x−35∘=45∘+n⋅180∘
Step 4: Solve for x in the domain 0∘≤x<360∘.
x=45∘+35∘+n⋅180∘x=80∘+n⋅180∘
For n=0: x=80∘.
For n=1: x=80∘+180∘=260∘.
For other integer values of n, x will be outside the given domain.
The solutions are x=80∘,260∘.
What's next?
Was this helpful?
Need help with your own homework?
Get instant step-by-step solutions to any question. Free to start.
This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
ScanSolve AI Answer
Here are the solutions for the given trigonometric equations: 5a) Solve ( + 30^) = for 0^ < 360^. Step 1: Expand ( + 30^) using the compound angle formula (A+B) = A B - A B. ( + 30^) = 30^ - 30^ Substitute the known values 30^ = sqrt(3)2 and 30^ = (1)/(2): ( + 30^) = sqrt(3)2 - (1)/(2) Step 2: Substitute this back into the original equation and rearrange to solve for . sqrt(3)2 - (1)/(2) = sqrt(3)2 = + (1)/(2) sqrt(3)2 = (3)/(2) Divide both sides by and (3)/(2): sqrt(3)3 = ()/() = sqrt(3)3 Step 3: Find the reference angle and solutions in the given domain. The reference angle for = sqrt(3)3 is 30^. Since is positive, lies in Quadrant I and Quadrant III. In Quadrant I: = 30^. In Quadrant III: = 180^ + 30^ = 210^. The solutions are = 30^, 210^. 5b) Solve sqrt(2)( - 45^) = 4 for 0^ < 360^. Step 1: Expand ( - 45^) using the compound angle formula (A-B) = A B + A B. ( - 45^) = 45^ + 45^ Substitute the known values 45^ = sqrt(2)2 and 45^ = sqrt(2)2: ( - 45^) = sqrt(2)2 + sqrt(2)2 Step 2: Substitute this back into the original equation and simplify. sqrt(2)(sqrt(2)2 + sqrt(2)2) = 4 ((2)/(2) + (2)/(2)) = 4 + = 4 Step 3: Rearrange the equation to solve for . = 4 - = 3 Divide both sides by : 1 = 3()/() = (1)/(3) Step 4: Find the reference angle and solutions in the given domain. The reference angle for = (1)/(3) is ((1)/(3)) ≈ 18.43^. Since is positive, lies in Quadrant I and Quadrant III. In Quadrant I: ≈ 18.43^. In Quadrant III: = 180^ + 18.43^ = 198.43^. The solutions are ≈ 18.43^, 198.43^. 6a) Solve 50^ x + 50^ x = 1 for 0^ x < 360^. Step 1: Recognize the compound angle formula (A+B) = A B + A B. The left side of the equation matches this form with A = 50^ and B = x. So, the equation can be rewritten as: (50^ + x) = 1 Step 2: Find the general solution for the angle (50^ + x). The sine function is equal to 1 when the angle is 90^ plus any multiple of 360^. 50^ + x = 90^ + n · 360^ Step 3: Solve for x in the domain 0^ x < 360^. x = 90^ - 50^ + n · 360^ x = 40^ + n · 360^ For n=0: x = 40^. For other integer values of n, x will be outside the given domain. The solution is x = 40^. 6b) Solve x - 35^ = 1 + x 35^ for 0^ x < 360^. Step 1: Rearrange the equation to match the compound angle formula for (A-B) = ( A - B)/(1 + A B). Divide both sides by (1 + x 35^): ( x - 35^)/(1 + x 35^) = 1 Step 2: Recognize the left side as (x - 35^). So, the equation becomes: (x - 35^) = 1 Step 3: Find the general solution for the angle (x - 35^). The tangent function is equal to 1 when the angle is 45^ plus any multiple of 180^. x - 35^ = 45^ + n · 180^ Step 4: Solve for x in the domain 0^ x < 360^. x = 45^ + 35^ + n · 180^ x = 80^ + n · 180^ For n=0: x = 80^. For n=1: x = 80^ + 180^ = 260^. For other integer values of n, x will be outside the given domain. The solutions are x = 80^, 260^. What's next?
