Section 1.6 : Solving Trig Equations with Calculators, Part II
5. Find all the solutions to 7cos2(3x)−cos(3x)=0. Use at least 4 decimal places in your work.
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Notice that we can factor a cosine out of each term to get,
cos(3x)(7cos(3x)−1)=0Now, we have a product of two factors that equals zero and so by basic algebraic properties we know that we must have,
cos(3x)=0OR7cos(3x)−1=0Each of these equations are similar to equations solved in the previous section. Therefore, we will be assuming that you can recall the solution process for each and we will not be putting in as many details. If you are unsure of the process you should go back to the previous section and work some of the problems there before proceeding with the solution to this problem.
We’ll start with,
cos(3x)=0From a unit circle we can see that we must have,
3x=π2+2πnOR3x=3π2+2πnn=0,±1,±2,…Notice that we can further reduce this down to,
3x=π2+πnn=0,±1,±2,…Finally, the solutions from this equation are,
x=π6+πn3n=0,±1,±2,…The second equation will take a little more (but not much more) work. First, isolating the cosine gives,
cos(3x)=17Using our calculator we get,
3x=cos−1(17)=1.4274From a quick look at a unit circle we know that the second angle in the range [0,2π] will be 2π−1.4274=4.8558.
Finally, the solutions to this equation are,
3x=1.4274+2πnOR3x=4.8558+2πnn=0,±1,±2,…x=0.4758+2πn3ORx=1.6186+2πn3n=0,±1,±2,…Putting all of this together gives the following set of solutions.
x=π6+πn3,x=0.4758+2πn3,ORx=1.6186+2πn3n=0,±1,±2,…If an interval had been given we would next proceed with plugging in values of n to determine which solutions fall in that interval. Since we were not given an interval this is as far as we can go.
Note that depending upon the amount of decimals you used here your answers may vary slightly from these due to round off error. Any differences should be slight and only appear around the 4th decimal place or so however.